LIPRARY 

UNIVERSITY  Of 
V      CALIFORNIA 


PLANE   AND    SPHERICAL 
TRIGONOMETRY 


THE   MARSH   AND  ASHTON    MATHEMATICAL  SERIES. 

BY 

WALTER   R.    MARSH, 
HEAD  MASTER  PINGRY  SCHOOL,   ELIZABETH,   N.J. 

AND 

CHARLES   H.   ASHTON. 
INSTRUCTOR  IN  MATHEMATICS,    HARVARD    UNIVERSITY. 


The  series  will  include  text-books  in 

* 

ELEMENTARY  ALGEBRA, 

COLLEGE  ALGEBRA, 

PLANE  AND  SOLID  GEOMETRY, 

PLANE  AND  SPHERICAL  TRIGONOMETRY, 

PLANE  AND  SOLID  ANALYTIC  GEOMETRY. 


PLANE    AND    SPHERICAL 
TRIGONOMETRY 

AN  ELEMENTARY  TEXT-BOOK 


BY 

CHARLES    H.   ASHTON,  A.M. 

ASSISTANT    PROFESSOR    OF    MATHEMATICS    IN    THE    UNIVERSITY    OF    KANSAS 

AND 

WALTER   R.    MARSH,  A.B. 

HEAD  MASTER   PINGRY   SCHOOL,   ELIZABETH,   N.J. 


NEW   YORK 

CHARLES   SCRIBNER'S   SONS 
1908 


COPYRIGHT,    1902,   BY 
CHARLES   SCRIBNER'S   SONS 


PREFACE 


THERE  have  been  two  distinct  methods  of  presenting 
the  subject  of  Trigonometry.  In  one  the  treatment  is 
purely  geometrical,  no  attention  being  paid  to  the  direc- 
tion of  lines.  Here  the  algebraic  signs  of  the  functions 
are  assigned  in  a  purely  arbitrary  manner,  and  none  of 
the  proofs  hold  except  for  the  particular  figure.  When 
this  method  is  employed,  the  result  must  be  shown  to  be 
general  by  some  algebraic  process. 

In  the  second  method  all  lines  have  direction  as  well  as 
magnitude,  and  the  proofs  are  given  in  such  a  form  that 
they  are  general  and  hold  for  every  possible  figure.  This 
would  seem  to  be  the  logical  method  of  developing  the 
subject ;  for  Trigonometry  is  the  connecting  link  between 
elementary  Geometry  and  those  subjects  in  which  Algebra 
and  Geometry  are  combined  in  such  a  way  that  the 
directed  line  must  be  used  constantly.  This  second 
method  has  been  employed  in  this  work,  which  is  in- 
tended as  a  text-book  for  a  fifty  hour  course  in  high 
schools  and  the  ordinary  first  year  classes  in  college,  and 
is,  therefore,  made  as  elementary  as  possible.  All  matter 
not  required  for  such  a  course  has  been  excluded. 

We  have  attempted  to  avoid  the  usual  mistake  of  mix- 
ing the  two  methods  mentioned  above.  Two  distinct 
proofs  of  the  Addition  Theorem  are  given.  The  first 
employs  the  method  of  projection,  the  formulas  for  which 
have  been  simplified  by  employing  the  French  symbol  for 


yi  PREFACE 

an  angle  between  two  directed  lines;  the  second  is  a 
geometric  proof,  similar  to  the  one  usually  given,  which 
has  also  been  made  perfectly  general.  The  second  dem- 
onstration may  be  preferred  in  a  shorter  course. 

Carefully  selected  exercises  are  given  at  the  end  of 
almost  every  article.  In  solving  triangles,  the  natural 
functions  are  used  where  the  solution  may  be  obtained 
easily  without  the  aid  of  logarithms.  This  method  has 
been  followed  because,  where  logarithms  are  used  exclu- 
sively, the  student  often  does  not  know  the  meaning  of 
the  operation  he  is  performing.  The  subject  of  trigono- 
metric equations,  which  is  usually  accorded  little  atten- 
tion, is  here  given  in  a  separate  chapter,  for  it  is  thought 
that  in  no  other  way  can  the  student  acquire  so  good  a 
knowledge  of  general  principles  or  so  great  skill  in 
applying  the  formulas. 

AUGUST,  1902, 


CONTENTS 

PAET   I 
PLANE   TRIGONOMETRY 

CHAPTER  I 
FUNCTIONS  OF  ACUTE  ANGLES 

ART.  PAGE 

1.  Definitions 1 

2.  Functions  of  complementary  angles 5 

3.  Variation  of  the  functions  as  the  angle  varies        .         .         .6 

4.  Functions  of  45°,  30°,  and  60° 7 

5.  Relations  between  the  functions     ......        9 

6.  Trigonometric  identities         .......      13 

CHAPTER  II 
RIGHT  TRIANGLES 

7.  Solution  of  right  triangles 16 

8.  Solution  by  the  aid  of  logarithms 18 

9.  Heights  and  distances 19 

CHAPTER  III 
FUNCTIONS  OF  ANY  ANGLE 

10.  Directed  lines 24 

11.  Theorem 25 

12.  Angles 26 

13.  The  measure  of  angles 27 

14.  Addition  of  directed  angles    .         .        .        .  '  .        .29 

vii 


viii  CONTENTS 

ART.  PAGE 

15.  Angles  between  directed  lines         ......  30 

16.  Functions  of  angles  of  any  magnitude  .....  31 

17.  Algebraic  signs  of  the  functions     : 33 

18.  Functions  of  the  quadrantal  angles 35 

19.  Line  values  of  the  functions 38 

20.  Variations  of  the  functions 39 


CHAPTER  IV 
RELATIONS  BETWEEN  THE  FUNCTIONS 

21.  Relations  between  the  functions  of  any  angle        ...  45 

22.  Functions  of  —  a,  -  ±  a,  TT  ±  a,  -  TT  ±  a           .         .         .         .  46 

23.  Reduction  of  the  functions  of  any  angle  to  functions  of  an 

angle  less  than  —........  50 

24.  Projection        .                 52 

25.  Projection  of  a  broken  line 53 

26.  Projections  on  the  axes  of  any  line  through  the  origin          .  54 

27.  Projections  of  any  line  on  the  axes    .     .         .        .         .         .54 

28.  Functions  of  the  sum  and  difference  of  two  angles        .         .  56 

29.  Second  method  of  finding  sin  (a  +  /?)  and  cos  (a  +  (3)          .  59 

30.  Tan  (a  ±  0) 62 

31.  Functions  of  twice  an  angle 64 

32.  Functions  of  half  an  angle 66 

33.  Sum  and  difference  of  the  sines  and  of  the  cosines  of  two 

angles 69 


CHAPTER  V 
INVERSE  FUNCTIONS  AND  TRIGONOMETRIC  EQUATIONS 

34.  General  values         .         .   - 72 

35.  Inverse  trigonometric  functions 76 

36.  Solution  of  trigonometric  equations 79 

37.  Trigonometric  equations 81 

38.  Trigonometric  equations 83 


CONTENTS  ix 

CHAPTER  VI 
OBLIQUE  TRIANGLES 

ART.  PAGB 

39.  Introduction    .         . 89 

40.  Law  of  the  sines      .         .         .         . ' 89 

41.  Law  of  the  cosines  .........  90 

42.  Law  of  the  tangents        ........  92 

43.  General  form  of  the  law  of  the  sines  and  the  law  of  the 

cosines ' 92 

44.  Solution  of  oblique  triangles .         .         .         .         .         .         .95 

45.  Case  I.     Given  two  angles  and  one  side         .         .         .         .95 

46.  Case  II.     Given  two  sides  and  the  included  angle         .         .  97 

47.  Solution  by  logarithms   .         . 98 

48.  Case  III.     Given  two  sides  and  the  angle  opposite  one  of 

them 99 

49.  Case  IV.     Given  the  three  sides     .         .         .         .         .         .103 

50.  Solution  by  logarithms   . 104 

51.  Area  of  an  oblique  triangle 107 

PART   II 
SPHERICAL   TRIGONOMETRY 

CHAPTER   VII 
RIGHT  AND  QUADRANTAL  TRIANGLES 

52.  Introduction 117 

53.  Right  triangles 119 

54.  Napier's  rules  of  circular  parts        .         .        .        .        .        .  122 

55.  Solution  of  right  triangles 123 

56.  Quadrantal  triangles 126 

CHAPTER  VIII 
OBLIQUE  TRIANGLES 

57.  Law  of  the  sines     .        .        .        .  ' 128 

58.  Law  of  the  cosines  129 


X  CONTENTS 

ART.  PAGE 

59.  Transformation  of  formulas 130 

60.  Napier's  analogies  .........  132 

61.  Solution  of  oblique  spherical  triangles 134 

62.  Case  I.     Given  the  three  sides 135 

63.  Case  II.     Given  the  three  angles 136 

64.  Case  III.     Given  two  sides  and  the  included  angle        .         .  137 

65.  Case  IV.     Given  two  angles  and  the  included  side        .         .  138 

66.  Case  V.     Given  two  sides  and  the  angle  opposite  one  of 

them 139 

67.  Case  VI.     Given  two  angles  and  the  side  opposite  one  of 

them    .  141 


PLANE   AND    SPHERICAL 
TRIGONOMETRY 


PART   I 
PLANE   TRIGONOMETRY 


CHAPTER   I 
FUNCTIONS   OF  ACUTE  ANGLES 

1.  Definitions.  — Let  ABO  (Fig.  1)  be  a  right  triangle, 
right-angled  at  C\  and  let  A'B' 0'  be  any  second  right 
triangle  which  is  similar  to  the  first. 

From  the  definition  of  similar  triangles  it  follows  that 

Z A  =  /.A',    and 

OB 
AB 


O'B'      AO     AC 


A'B'      AB     A'B' 


and 


OB      C'B' 


AO     AC' 


These  five  equations  between  the  angles  and  the  ratios 
of  the  sides  are  true  for  any  pair  of  similar  right  tri- 
angles. Bat  any  two 
right  triangles  are  sim- 
ilar, if  an  acute  angle  of 
the  one  is  equal  to  an 
acute  angle  of  the  other, 
or  if  any  pair  of  homolo- 
gous sides  are  propor- 
tional.  Hence  if  any 
one  of  these  five  equations  hold,  the  remaining  four 
also  hold  and  the  triangles  are  similar. 

1 


PLANE   TRIGONOMETRY 


[Cn.  I,  §  1 


It  follows  then  that  in  every  triangle  equiangular  with 
ABO  the  ratios  of  the  sides  are  the  same  as  in  ABO, 
and  that  in  every  right  triangle  not  equiangular  with 
ABO  these  ratios  are  not  the  same  as  in  ABO;  and, 
conversely,  if  any  right  triangle  has  any  one  of  these 
ratios  equal  to  the  corresponding  ratio  of  ABO,  the 
triangle  must  be  equiangular  with  ABO. 

The  ratio,  then,  of  any  two  sides  of  a  right  triangle  is 
a  number  which  is  entirely  independent  of  the  lengths  of 

those  sides,  and,  if  the 
angle  A  is  fixed,  each  of 
these  ratios  has  a  deter- 
minate value  which  is  dif- 
ferent from  the  value  of 
the  corresponding  ratio 

A          ^X*          O     A'  C'     for     any     other     angle. 

FlG>1'  These    three    ratios    of 

the  sides  of  the  triangle  may  then  be  spoken  of  as 
functions  of  either  of  the  acute  angles  of  the  triangle. 
(One  quantity,  y,  is  said  to  be  a  function  of  another, 
x,  when  y  has  a  determinate  value,  or  values,  for  every 
value  given  to  x.) 

It  has  been  found  convenient  to  give  names  to  these 
ratios  as  follows :  If  A  is  either  of  the  acute  angles  of  a 
right  triangle  (see  Fig.  1), 

sine  of  A  =  sin  . 


cosine 


tangent 


AB      hypotenuse 

£  A  AO     adiacent  leg 

of  A  =  cos  A  =  -  ^  =  —  J  -  -•> 
AB      hypotenuse 


of  A  =  ten  A=<£=  °p,posite  }eS 
AC     adjacent  leg 


CH.  I,  §  1]  FUNCTIONS   OF  ACUTE   ANGLES  3 

The  abbreviations  in  the  second  column  should  always 
be  read  the  same  as  the  first  column. 

Names  are  also  given  to  the  reciprocals  of  these  ratios 
as  follows  : 

,.    ,,  A      AB      hypotenuse 

cosecant    of  A  =  esc  A  =  •— —  =    ^    . — - — , 

CJD      opposite  leg 

f   A  A      AB      hypotenuse 

secant       of  A  =  see  A  =  -—  =  — ^ — * 

AC     adjacent  leg 

•  '.  A      AC     adjacent  leg" 

cotangent  of  A  =  cot  A  —  — —  =  — J — : — — =-. 

CB     opposite  leg 


From  these  definitions  it  will  be  seen  at  once  that 


esc  A  =  — ,    sec  A  = -,    and    cot  A  = 


sin  A  cos  A  tan  A 

The  last  three  ratios  are  used  much  less  frequently  than 
the  first  three,  and  will  usually  be  studied  as  the  recip- 
rocals of  the  first  set.  It  is  in  this  way  that  the  student 
is  advised  to  memorize  their  definitions. 

There  are  also  two  other  functions  which  are  occasion- 
ally used.     They  are  defined  by  the  equations, 

versed  sine  of  A  =     vers  A  =  1  —  cos  A. 

coversed  sine  of  A  =  covers  A  =  1  —  sin  A. 

» 

The  student  should  note  that  these  eight  quantities  are 
all  abstract  numbers^  since  they  are  the  ratios  of  two  lines. 
They  are  called  the  trigonometric  functions  of  the  angle  A. 


PLANE   TRIGONOMETRY 


[On.  I,  §  1 


EXERCISE    I 

In  the  following  problems,  C  is  the  right  angle  of  the 
right  triangle  ABC,  and  the  small  letters  a,  b,  c  are  used  to 
represent  the  lengths  of  the  sides  opposite  the  corresponding 
angles  j  that  is,  c  represents  the  hypotenuse,  a  and  b  the  legs. 

1.  Find  all  the  functions  of  the  angle  A,  if 

(a)  a=   5,     6  =  12.  (c)    a=   8,     c  =  17. 
(6)    6=   4,     c=   5.  (d)   a  =10,     c=15. 

2.  What  are  the  functions  of  B  in  the  same  triangles  ? 

3.  What  relations  do  you  notice  between  the  functions  of 
A  and  B  ? 

4.  Construct  the  angle  whose  (a)  sine  is  f, 

(b)  tangent  is  5,  ^y  (d)  cotangent  is  2, 

(c)  cosine  is  J,         -J  >>  Y  \  (e)  secant  is  3. 

5.  Draw,  on  a  large  sheet  of  paper,  a  line  AC  10  in.  long. 
At  C  erect  a  perpendicular,  and  at  A  lay  off  (with  a  protractor) 
angles  of  10°,  20°,  etc.      Measure  carefully  the  sides  of  the 
triangles  thus  formed,  and  from  these  measurements  determine 
(to  two  decimal  places)  the  functions  of  the  angles  given  in 
the  following  table  : 


A 

sin  A 

cos  A 

tan  A 

cot  A 

10° 

.1737 

.9848 

.1763 

5.6713 

20° 

.3420 

.9397 

.3640 

2.7475 

30° 

.5000 

.8660 

.5774 

1.7321 

40° 

.6428 

.7660 

.8391 

1.1918 

50° 

.7660 

.6428 

1.1918 

.8391 

60° 

.8660 

.5000 

1.7321 

.5774 

70° 

.9397 

.3420 

2.7475 

.3640 

80° 

.9848 

.1737 

5.6713 

.1763 

6.    By  the  aid  of  the  table  given  above  find  the  legs  of  a 
right  triangle,  if  A  =  40°  and  c  =  12. 


CH.  I,  §  2]     FUNCTIONS  OF  ACUTE  ANGLES  5 

SOLUTION.  —  By  definition, 

a  b 

sin  A  =  - .    and    cos  A  =  -  • 

c'  c 

Hence     a  =  c  sin  A  =  12  sin  40°  =  12  x  .6428  =  7.7136, 
and          b  =  c  cos  A  =  12  cos  40°  =  12  x  .7660  =  9.192. 

7.  Find  c,ifA  =  30°  and  a  =  8. 

8.  Find  A,iia  =  171  and  c  =  500. 

9.  Find  A,  a,  and  c,  if  B  =  50°  and  6  —  6. 

10.    Find  .4,  5,  and  c,  if  a  =  91  and  b  =  250. 

li 
<g«        11.    Find  5,  c,  and  a,  if  ^4  =  70°  and  b  =  10. 

12.    Find  A,  a,  and  6,  if  B  =  20°  and  c  =  20. 

2.  Functions  of  complementary  angles.  —  Using  the 
definitions  of  Art.  1  for  the  functions  of  angle  B  of  the 
triangle  ABC,  we  have  (see  Fig.  2) 


sin  B  =  ^  =  cos  A  =  cos  (90°  -  B), 
cos  B  =  — —  =  sin  A  =  sin  (90°  —  #), 
tan  B  =  ^  =  cot  A  =  cot  (90°  -  B), 
CQiB  =  ~  =  tan  J.  =  tan  (90°  -  5), 
sec  B  =  -—  =  csc  A  =  csc  (90°  —  B), 


FIG.  2. 

csc  B  =         =  sec  A  =  sec  (90°  -  5). 


6 


PLANE  TRIGONOMETRY 


[Cn.  I,  §  3 


From  which  it  appears  that  any  function  of  an  acute 
angle  is  equal  to  the  co-named  function  of  its  complementary 
angle. 

EXERCISE  II 

1.  Express  as  functions  of  angles  less  than  45°: 

sin  70°,   cos85°,   sec  63°,   tan  56°  48',   cot  89°  56 

2.  If  cos  x  =  sin  x,  find  a  value  of  x. 

3.  If  tan  x  =  cot  2  x,  find  a  value  of  x. 

4.  If  cos  x  =  sin  (45°  +  2  x),  find  a  value  of  x. 

5.  If  A,  B,  and  C  are  the  angles  of  a  triangle,  show  that 
cos A=.  si 


3.  Variation  of  the  functions  as  the  angle  varies.  —  Since 
the  trigonometric  functions  have  been  defined  by  the  aid 
of  a  right  triangle,  the  angle  A  must  be  between  0° 
and  90°,  not  including  either  of  these.  Later  we  shall 
give  definitions  of  the  functions  which  will  include  all 
angles,  but  for  the  present  we  shall  confine  our  discus- 
sion to  angles  between  0°  and 
90°. 

Let   the   angle  A   begin  with 
small  values  and  increase  toward 
B    90°,   and    consider    the    changes 
which  will  occur  in  each  func- 
tion.    To  fix  our  ideas,  let  AC 
remain  the  same,  while  AB  and 
CB    vary.       For    small    angles 
OB    is    small,    and    AB    nearly 
FIG.  3.  equal  to  A  0.      Hence  the  sine 


CH.  I,  §  4]  FUNCTIONS  OF  ACUTE   ANGLES 

frl)  and  the  tan^ent  (^]  are  small>  the  cosine 


.AB) 

nearly  unity,  the  secant  (reciprocal  of  the  cosine)  a  little 
more  than  unity,  while  the  cotangent  and  cosecant  (recip- 
rocals of  the  tangent  and  sine)  are  very  large. 

As  the  angle  increases,  both  BC  and  AB  increase  and 
approach  equality.  Hence  the  sine,  tangent,  and  secant 
increase,  while  the  cosine,  cotangent,  and  cosecant  de- 
crease. As  the  angle  A  approaches  90°,  AB  and  CB  are 
very  large  and  nearly  equal.  Hence  the  sine  approaches 
unity,  and  the  tangent  and  secant  increase  indefinitely. 
The  cosecant  decreases  toward  unity,  while  the  cosine  and 
cotangent  decrease  indefinitely. 

It  appears,  then,  that  for  angles  less  than  90°  the 

sine  and  cosine  are  less  than  unity, 
secant  and  cosecant  are  greater  than  unity, 
tangent  and  cotangent  may  have  any  positive  value. 

4.    Functions  of  45°,  30°,  and  60°. 

(a)  Functions  of  45°. 

In  the  right  triangle  ABO  (Fig.  4)  let  A  O=  OB  =  a. 
Then  angle  A  =  angle  B= 45°,  and  AB  =  Vo"2  +  a2  =  a  V2~. 

It  follows  at  once  from  the  defini- 
tions of  the  functions  that 

sin  45°  =  cos  45°  =  -?—  =  \  V2, 


tan  45°  =  cot  45°  =    -    =  1, 
a 

sec  45°  =  esc  45°  =  —  =  V2. 


PLANE   TRIGONOMETRY 


[Cii.  I,  §  4 


(b)  Functions  of  30°  and  60°. 

Let  each  side  of  the  equilateral  triangle  ABD  (Fig.  5) 

be  represented  by  2  a.  Bisect 
the  angle  B  by  the  line  BC. 
Then  this  line  bisects  the  base 
AD  and  is  perpendicular  to  it. 
A  right  triangle  AB  C  is  thus 
formed,  in  which  angle  A  =  60° 
and  angle  £=30°.  Also  AB= 2  a, 
AC=  a,  and  BC=  V-4  a2-  a2  = 
a  VS.  It  follows  at  once  from  the 
definitions  of  the  functions  that 


D 


FIG.  5. 


sin  60°  =  cos  30°  = 


V3 

9   ' 


cos  60°  =  sin  30°  =  ^-   =  |, 


tan  60°  =  cot  30° 

a 

cot  60°  =  tan  30°  =  --^  =  i  VS, 
sec  60°  =  esc  30°  = 
esc  60°  =  sec  30°  = 


EXERCISE  III 

Find  the  numerical  value  of 

1.    2  sin  30°  cos  30°  cot  60°.  2. 


tan2  60°  +  2  tan2  45°. 


NOTE.  —  tan2  60°  is  equivalent  to  (tan  GO0)2,  or  the  square  of  the 
tangent  of  60°. 


CH.  I,  §  5]  FUNCTIONS   OF  ACUTE   ANGLES 

3.  tan3  45°  +  4  cos3  60°. 

4.  4  cos2  45°  +  tan2  60°  +  3  sec2  30° 

5.  sin  60°  cos  30°  +  cos  60°  sin  30° 

6.  sin2  45°  +  cos2  60°  -  sin2  30°. 

Prove  tluit 

PAO         2  tan  30° 
= 


8.  sin  60°  =  2  sin  30°  cos  30°. 

9.  cos  60°  =  1  -2  sin2  30°. 

10.    cos2  60°  :  cos2  45°  :  cos2  30°  =  1  :  2  :  3. 

5.  Relations  between  the  functions.  —  It  appeared  at 
once  from  the  definitions  of  the  six  trigonometric  func- 
tions that  they  were  not  all  independent  ;  for  three  of 
them  were  defined  as  the  reciprocals  of  the  other  three. 
From  these  definitions  we  have  the  following  relations  : 

sin  A  esc  A  =  19  [1] 

cos  .4  sec  .4  =  1,  [2] 


*     hi 

s         /I'1 

We  shall  now  show  that  there  are  other  relations  be- 
tween these  functions ;  that,  intact,  each  depends  on  the 
others,  so  that,  if  one  is  giveii,  all  the 
others  may  be  found. 

In  the  right  triangle  ABO, 


Dividing  by  Alf, 
we  have 


A& 
or  siii2^!  +  cos2^  =  1.  [4]  FIG.  G. 


10  PLANE  TRIGONOMETRY  [Cn.  I,  §  5 

Again  dividing  the  same  equation  by  AC2, 


we  have  -        +  1  ,= 

AC'2 

or  1  +  tan2  A  =  sec2  A.  [5] 

Dividing  the  same  equation  by 
wehave  1 


or  l+cot2^  =  csc2^.  [6] 

Dividing  sin  A  by  cos  A,  we  have 

sin  A     ~AE      CB 


AB 

But  tanJ.  =  — 

A  C 


Hence 

Since  cot  A  = 


cos^l 


tan^l 
cot  A=^.  [8] 


These  formulas  should  be  carefully  memorized  by  the 
student,  as  they  are  in  constant  use  throughout  the 
study  of  the  subject.  The  student  should  also  be  per- 
fectly familiar  with  the  following  forms,  which  may  be 
obtained  easily  from  the  formulas  given  above  : 


CH.  I,  §  5]  FUNCTIONS  OF  ACUTE  ANGLES  11 


sin.  A  =  VI  —  cos2  A.  cos  A  =  Vl  •—  sin2^., 


sec  A  =  Vl  -h  tan2  A,  esc  A  =  Vl  +  cot2  A, 


tan  J.  =  Vsec2  A  —  1,  cot  J.  =  VcscM  —  1. 

One  of  the  simplest  applications  of  these  formulas  is  in 
finding  the  remaining  functions,  when  any  one  function 
of  an  angle  is  given. 

For  example,  let  sin  A  =  |-. 


Then  cos  A  =  Vl-sin2J.  =  Vl  -  J  =  -|  V3. 


cos^L      1V3 


cot  ^1  =  __  =  -^  =  V3, 
tan^l 


sec  ^  = 


=  -^  =  *  V3, 


cosJ.      1.V3 


i 

esc  A  =  -  -  =  -  =  2. 
sin  A 


Again,  let  tan  A  =  5. 
Then  cot  ^4  = 


sec  ^4.  =  Vl  +  tan2  A  =  V26, 


in^  =  _J_=  5  V26, 
esc  ^4. 


sec  A 


12  PLANE  TRIGONOMETRY  [Cn.  I,  §  5 

EXERCISE  IV 
Find  all  the  other  functions  of  A,  if 

1.  sinJ.  =  f.  6.  esc  ^4  =  10. 

2.  cos  A  =  i.  7.  sin -4  = -J-}. 

3.  tan  J.  =  3.  8.  cos  A  =  f. 

4.  cot  A  =  6.  9.  tan  ^4  =  -|. 

5.  sec  A  =  2.  10.  sec  A  =  a. 

11.  Find  all  the  other  functions  of  45°  from  the  fact  that 
tan  45°  =  1. 

12.  Find  all  the  other  functions  of  GO0  from  the  fact  that 
sec  60°  =  2. 

13.  Find  all  the  other  functions  of  30°  from  the  fact  that 
cot30°  =  V3. 

14.  Find  all  the  other  functions  of  A  in  terms  of 

(a)    sin  A,  (b)   cos  A,          (c)    tan  A,          (d)    sec  A. 

15.  Transform  each  of  the  following  expressions  into  an- 
other form  which  shall  contain  sin  A  only : 

(a)   ^AeofA+^A cot  A      . 

cos  A     cos  A  sm  A 

cos2  A     cot  A 
(c)    sec2  A  +  cos2  A  -  tanl4  cot2  A. 

v       16.    Transform  each  of  the  following  expressions  into  an- 
other form  which  shall  contain  cos  A  only : 

(a)   sin  A  cos  A  tan  A  cot  A. 

(b}    sec  A  —      1     4.  tan  A 
cos  A      sin  A 

(c)    sin2  A  +  cos2  A  —  tan2  A  —  cot2  -4. 


.  CH.  I,  §  6]     FUNCTIONS  OF  ACUTE  ANGLES          13 

17.  Transform  each  of  the  following  expressions  into  an- 
other form  which  shall  contain  tan  A  only : 

(a)  sin  A  cos  A  +  cos  A  tan  A  -f  sin  A  cot  A. 

/7N      •      A  A      1  —  sin  A  cos  A 

(b)  sin  A  cot  A — 

1  +  sin  A  cos  A 

(c)  sec  A  —  esc  A  +  cot  A. 

6.  Trigonometric  identities. — Another  important  use  of 
the  formulas  obtained  in  Art.  5  is  in  proving  the  identity 
of  certain  trigonometric  expressions. 

The  student  should  here  make  himself  familiar  with  the 
distinction  between  identities  and  conditional  equations, 
such  as  usually  occur  in  algebra.  An  identity  is  an  equa- 
tion in  which  the  two  members  are  equal  for  every  possible 
value  of  the-  variables  in  it ;  while  a  conditional  equation 
holds  only  for  certain  values  of  these  variables.  For 
example,  2x  +  3  x  =  5  x  and'  sin2  x  +  cos2  x  —  1  are  identi- 
ties ;  while  x2  —  2  x  =  5  and  sin  x  —  cos  x  are  conditional 
equations. 

We  are  at  present  concerned  only  with  the  proof  of 
identities.  The  method  to  be  employed  is  to  change  the 
form  of  one  of  the  members  of  the  equation,  by  the  applica- 
tion of  the  formulas  of  Art.  5,  until  it  has  been  made  to 
assume  the  form  of  the  other  member.  Skill  in  choosing 
the  proper  formulas  to  accomplish  this  result  with  the  least 
labor  is  acquired  only  after  considerable  practice  ;  but  the 
form  which  we  wish  to  obtain  will  soon  suggest  to  the 
student  which  formulas  he  should  use. 

The  following  examples  illustrate  the  mode  of  pro- 
cedure : 


14  PLANE   TRIGONOMETRY  [Cn.  I,  §  6 

EXAMPLE  1.     Prove  that  (1  —  cos2  A)  sec2  A  =  tan2  A. 
If  we  replace  1  —  cos2  A  by  its  equal  sin2^L,  and  sec2  A 

1  2     A 

by  — — -,  the  first  member  becomes  S1I\  J ,  which  is  evi- 
cos^-4  cos2JL 

dently  equal  to  tan2  A. 

EXAMPLE  2.     Prove  that  esc2  A  tan2  A  -  1  =  tan2  A. 
Here  it  is  probably  simplest  to  express  the  first  member 
in  terms  of  the  sine  and  cosine.     This  is  often  advisable. 
It  becomes 

1        sin2J.      i  1  2/1 

—  1  =  — _ —  —  1  =  sec2  A  —  1  =  tan2  A. 


sin2  A    cos2  A  cos2  A 

EXAMPLE  3.     Prove  that  sec  A  —  tan  A  sin  A  =  cos  A. 
As  in  the  previous  example,  express  the  first  member  in 
terms  of  the  sine  and  cosine.     It  becomes 

1      _  sin  A  .      M  _  1  —  sin2  A  _  cos2  A  _          . 
cos  A     cos  A  co§  A          cos  A 

EXERCISE  V 
Prove  the  following  identities  : 

1 .  cos  A  tan  A  =  sin  A. 

2.  sin  A  sec  A  =  tan  A. 

3.  cos  A  esc  A  =  cot  A. 

4.  sin  A  sec  A  cot  ^1  =  1. 

5.  sin2  ^4  sec2  ^4  =  sec?  A  —  1. 

^     sin  J.      cos^l  _  ^ 
esc  A     sec  yl 

7. 


COS^ 


CH.  I,  §  0]     FUNCTIONS  OF  ACUTE  ANGLES          15 

8.  tan2  ^4  —  sin2  .4  =  sin4  .4  see2  .4. 

9.  cot2  A  —  cos2  A  =  cot2  A  cos2  A. 

# 

10.  tan  A  +  cot  A  =  sec  A  esc  A. 

11.  sin4  A  —  cos4  .4  =  sin2  A  —  cos2  .4. 

12. 


cos  A         sin  A 
tan  A  +  cot  J 


13.    -  =  sin  A  cos  A. 


14 


15.  sin.4 


1  —  tan  A     1  —  cot  A 
16.    sec2  A  esc2  .4  =  tan2  A-\-  cot2  .4  +  2. 

'      17.    sin  A  (tan  A  —  1  )  —  cos  A  (cot  ^4  —  1)  =  sec  A  —  esc  A. 
^  18. 

V  19.  ^ 

1  +  sin  A 


20.  (sin  A  +  cos  ^4)  (tan  A  +  cot  ^4)  =  sec  A  +  esc  A 

2  1  .  tan2  yl  —  cot2  A  =  sec2  A  esc2  J.  (sin2  J.  —  cos2  A)  . 

22.  (1  +  cot  ^4  —  esc  ^4)  (1  +  tan  J.  +  sec  A)  =  2. 

23.  2  vers  ^4  +  cos2  A  =  1  +  vers2  A. 

24.  cot4  ^4+  cot2  A  =  esc4  ^4  -cse2  A. 

25  .  (sin  ^4  +  esc  ^4)2  +  (cos  A  -f  sec  ^4)2  =  tan2  A  +  cot2  ^1  +  7 


CHAPTER   II 


RIGHT   TRIANGLES 

7.    Solution  of  right  triangles.  —  A  right  triangle  may 
be   solved  (that  is,  the   unknown  parts  may  be  found) 
when  any  two  parts,  one  of  which  is  a  side,  are  given. 
It  is  convenient  to  letter  the  triangle  as  in  Fig.  7,  the 

small  letters  a,  &,  c  representing 
the  lengths  of  the  sides  opposite 
the  corresponding  angles. 

When  either  A  or  B  is  known, 
the  other  angle  may  be  deter- 
mined by  subtracting  the  given 
angle  from  90°.  When  any  two 
of  the  sides  are  known,  the  third 

side  may  be  found  by  the  aid  of  the  equation  a2  +  b2  =  ft. 
But  when  it  is  necessary  to  find  the  angles,  having  given 
two  sides,  or  to  find  the  other  sides,  having  given  one  side 
and  an  angle,  the  trigonometric  functions  must  be  intro- 
duced. The  unknown  parts  may  then  be  found  by  the 
aid  of  the  definitions  of  the  functions, 


v      b 
FIG.  7. 


sin  A  = 

cos  1 

?  =  i 

C 

J 

cos  A  = 

sin  1 

c 

# 

tanA  = 

cot  J 

5=6 

16 

CH.  II,  §  7]  RIGHT  TRIANGLES  17 

Choose  that  one  of  these  formulas  tvhicJi  contains  the  two 
given  parts  and  one  unknown  part,  and  solve  for  the  un- 
known part;  its  numerical  value  may  then  be  found  by 
the  aid  of  a  table  of  trigonometric  functions. 

If  necessary,  repeat  the  operation  with  one  of  the 
other  formulas  to  find  the  remaining  parts. 

Each  of  the  unknown  parts  should  be  determined 
directly  from  the  given  parts,  without  using  the  results 
of  a  previous  operation.  The  accuracy  of  the  work  may 
then  be  tested  by  determining  one  of  the  given  parts 
from  the  parts  just  found. 

The  different  cases  which  may  arise  might  be  separated, 
and  the  particular  formulas  to  be  used  in  each  indicated  ; 
but  it  is  thought  best  to  let  the  student  determine  the 
best  method  of  solving  each  problem,  only  giving  a  few 
typical  examples. 

EXERCISE  VI 

2.    Given  c  =  16  and  A  =  26°  15' ;  find  the  remaining  parts. 
To  obtain  a,  use 

sm^.  =  -,   or   a  =  c  sin  A  =  16  x  .44229  =  7.0766. 

c 

To  obtain  b,  use 

cos^l  =  -,    or   b  =  c  cos  A  =  16  x  .89687  =  14.35. 

C 

The  accuracy  of  these  results  may  be  tested  by  determining 
whether  they  satisfy  the  equation  a2  -f  b2  =  c2. 
The  angle  B  is  the  complement  of  A,  or  63°  45'. 

2.    Given  a  =  8  and  b  =  12 ;  find  the  remaining  parts. 
Here  c  may  be  found  at  once  from  the  equation  c2  =  a2  +  b'2. 
From  this  equation,  c  =  V64  +  144  =  14.44. 


18  PLANE   TRIGONOMETRY  [Cn.  II,  §  8 

To  obtain  A,  use  tan  A  =  -  =  —  =  .66666. 

6      12 

From  this,  by  the  aid  of  the  tables, 

A  =  33°  £1'.     B  =  90°  —  A  =  56*  19'. 

These  results  may  be  tested  by  determining  whether  they 
satisfy  either  the  equation  for  sin  A  or  cos  A. 

Find  the  remaining  parts  in  each  of  the  following  problems  : 
3.    a  =  3,     6  =  4.  7.    6  =  13,  c  =85. 

S    4.    a  =  9,     6  =  40.  y    8.    a  =  12,  A  =  34°  42'. 

5.    a  =  12,  c  =  13.  ^9.    c=15,  £  =  23°  34'. 

I.   a  =  60,  c  =  61.  /xlO.    6  =  17,  A  =  13°  52'. 


8.  Solution  by  the  aid  of  logarithms.  —  The  student 
should  now  make  himself  familiar  with  the  use  of  loga- 
rithms by  reading  the  explanation  of  the  tables,  and  by 
doing  some  of  the  problems  given  there.  Logarithms  are 
not  of  great  use  in  solving  right  triangles,  since  they  do 
not  greatly  decrease  the  labor  involved ;  but  it  is  best  to 
become  familiar  with  their  use  in  solving  simple  problems. 

EXERCISE  VII 
1.    Given  a  =  12.73  and  c  =  43.18  ;  find  the  remaining  parts. 

sin  A  =  -  -  sin  B  =  -,  or  6  —  c  sin  B. 

c  c 

Hence  log  sin  A= log  a— log  c.     Hence  log6  =  logc-J-logsiiiJ3. 

log  a  =  1.10483  log  c  =  1.63528 

log  c  =  1.63528  log  sin  £  =  9.98026 

log  sin  A  =  9.46955  log  6  =  1.61554 

.4  =  17°   8' 47".  6  =  41.26. 

£  =  90° -^1  =  72°  51 '13". 


CH.  II,  §  9] 


RIGHT  TRIANGLES 


19 


2.    Given  ^4=43° 48'  and  I  =67.92  ;  find  the  remaining  parts. 


A 

cos  A  =  -,  or  c  =  —    —  • 
c  cos  A 


tan -4  =      or  a=btanA. 


Hence  log  c = log  b  —  log  cos  A.    Hence  log  a = log  b  -f-  log  tan  A. 

log  6  =  1.83200  log  b  =  1.83200 

log  cos  A  =  9.85839  log  tan  A  =  9.98180 

log  c  =  1.97361  log  a  =  1.81380 

c  =  94.10.  a  =  65.13. 

Find  the  remaining  parts  in  each  of  the  following  problems : 


3.  «  =  5678,  6  =  6789. 

'  4.  a  =  2222,  c  =  3333. 

5.  a  =  .4545,  c=.5454. 

,  6.  «  =  4567,  4  =  23°  52'. 

7.  c  =  . 8765,  £  =  27° 25'. 

8.  a  =  206.14,  .4  =  24°  24'. 

9.  c  =  2.383,  5  =  32°  42'. 

10.  a  =  1758,  6  =  1312.7. 

11.  a  =  .581,  6  =  13.5. 


12.  6  =  75.84,  A  =  S7°32'. 

13.  6  =  8.4,  c  =  14. 

14.  a  =  795,  6  =  164. 

15.  c=543.3,5=17°20'35". 

16.  a  =  1.456,  yl=3°26'42". 

17.  a  =  .0065,  c  =  .0094. 

18.  c  =  765,  .4=84°  16'. 

19.  c  =  1000,^l  =  750. 

20.  a  =  1.006,  c  =  1.06. 


9.  Heights  and  distances.  —  One  of  the  applications  of 
Trigonometry  is  in  finding  the  height  of  objects  or  the 
distance  between  points,  when  that  height  or  distance  can- 
not be  easily  measured.  This  is  accomplished  by  measur- 
ing other  easily  accessible  lines  and  certain  angles,  and 
computing  the  desired  lines 
from  these  data. 

It  is  necessary  to  explain 
a  few  terms  used  in  such  prob- 
lems. Let  A  and  B  be  two 
points  not  in  the  same  horizon- 
tal  plane,  and  let  A  C  and  BD  FIG.  8. 


20  PLANE   TRIGONOMETRY  [Cn.  II,  §  9 

be  horizontal  lines  through  these  points.  If  B  is  observed 
from  J.,  the  angle  CAB  is  called  the  angle  of  elevation 
of  B  from  A.  If  A  is  observed  from  B,  the  angle  DBA 
is  called  the  angle  of  depression  of  A  from  B. 

Again,  let  AB  be  any  object, 
and  let  Q  be  the  point  of  ob- 
servation. The  angle  ACB  is 
spoken  of  as  the  angle  which  AB 
subtends  at  C. 

The  earlier  problems  of  the 
following  exercise  give  simple 

right  triangles,  and  can  be  solved  by  the  student  without 
further  assistance. 

EXERCISE  VIII 

1.  A  vertical  pole  13  ft.  high  casts  a  shadow  21  ft.  long. 
What  is  the  angle  of  elevation  of  the  sun  at  this  moment  ? 

2.  How  high  is  the   tower  which   casts  a   shadow  125  ft. 
long,  when  the  angle  of  elevation  of  the  sun  is  28°  20'  ? 

3.  What  is  the  angle  of  elevation  of  the  top  of   a  tower 
150  ft.  high  at  a  point  in  the  same  horizontal  plane  as  its  foot 
and  200  ft.  distant  ? 

4.  From  the  top  of  a  tower  100  ft.  high,  the  angle  of  de- 
pression of  an  object  in  the  same  horizontal  plane  as  its  foot 
is  found  to  be  31°.     How  far  is  the  object  from  the  tower  ? 

5.  The  angle  of   elevation  of  the  top  of  a  tower  from  a 
point  in  the  same  horizontal  plane  and  57  ft.  from  its  foot  is 
found  to  be  22°  14'.     How  high  is  the  tower  ? 

6.  A  building  95  ft.  high  stands  on  the  bank  of   a  river. 
The  angle  of  elevation  of  the  top  of  the  building  from  the 
opposite  bank  of  the  river  is  found  to  be  25°  10'.     Find  the 
breadth  of  the  river. 


CH.  II,  §  9J  EIGHT  TRIANGLES  21 

7.  The  two  equal  legs  of  an  isosceles  triangle  are  each 
10  ft.  and  the  opposite  side  is  6  ft.     Find  the  three  angles. 

NOTE.  —  Draw  the  altitude  of  the  triangle.  Then  there  will  be 
two  equal  right  triangles  formed,  in  which  the  base  is  3  ft.  and  the 
hypotenuse  is  10  ft. 

8.  From  a  point  directly  in  front  of  the  middle  of  a  build- 
ing and  100  ft.  distant,  the  length  of  the  building  is  found  to 
subtend  an  aogle  of  34°  15'.     How  long  is  the  building? 

9.  Two  trees  stand  directly  opposite  each  other  on  a  straight 
road  80  ft.  wide.     From  a  point  in  the  centre  of  the  road  the 
line  joining  their  trunks  subtends  an  angle  of  5°  28'.     How  far 
is  the  point  from  the  trees  ? 

10.  A  circular  balloon  10  yd.  in  diameter  is  noted  by  an 
observer  to  subtend  an  angle  of  40'.     At  the  same  time  the 
angle  of  elevation  of  its  apparent  lowest  point  is  50°  10'.    Find, 
approximately,  the  height  of  the  balloon. 

11.  A.  flagstaff  20  ft.  long  stands  on  the  corner  of  a  building 
150  ft.  high.     Find  the  angle  subtended  by  the  flagstaff  at  a 
point  100  ft.  from  the  foundation  of  the  corner. 

'  12.  A  strip  of  river  bank  is  straight.  It  is  300  ft.  long  and 
it  subtends  a  right  angle  at  a  point  on  the  opposite  shore.  The 
a;igle  between  a  line  drawn  from  the  point  to  one  end  of  the 
strip  and  the  perpendicular  from  the  point  to  the  strip  is  15°. 
Find  the  width  of  the  river. 

13.  A  ladder  30  ft.  long  leans  against  a  house  on  one  side 
of  a  street  making  an  angle  of  60°  with  the  street.  •  On  turning 
the  ladder  about  its  foot  till  the  top  touches  the  house  on  the 
opposite  side,  the  angle  is  found  to  be  30°.     Find  the  width  of 
the  street. 

14.  To  find  the  height  of  a  chimney  a  distance  of  125  ft. 
is  measured  from  its  base.     From  the  point  thus  reached  the 
angle  of  elevation  of  the  top  of  the  chimney  is  found  to  be 
48°  25'.     What  is  the  height  of  the  chimney  ? 


22 


PLANE  TRIGONOMETRY 


[Cn.  II,  §  9 


15.  From  the  top  of  a  telegraph  pole  35  ft.  tall  a  wire  50  ft. 
long  is  stretched  to  the  ground.  Find  the  angle  which  the 
wire  makes  with  the  ground. 

'16.  A  man  lies  on  the  ground  with  his  eye  to  the  edge  of  a 
well  and,  looking  into  the  well,  he  sees  the  reflection  of  the 
opposite  edge  in  the  water.  The  direction  in  which  he  looks 
makes  with  the  vertical  an  angle  of  17°.  The  well  is  6  ft. 
broad.  Find  the  distance  from  the  edge  of  the  well  to  the 
surface  of  the  water. 

NOTE.  —  The  angle  of  reflection  equals  the  angle  of  incidence. 

"   IT.    From  one  point  6t  oblservaQcm^cne  angle  of  elevation)  of 

the  top  of  a  building  is  found  to  be  35°.  The  observer  walks 
100  ft.  directly  away  from  the  building  in  the  same  horizontal 
plane  and  then  finds  the  angle  of  elevation  of  the  top  of  the 
building  to  be  25°.  Find  the  height  of  the  building. 

SOLUTION.  —  Let  AB  (Fig.  10)  represent   the   face   of   the 

building,  and  let  C  be  the 
first  point  of  observation 
and  D  the  second  point. 
Then  .405  =  35°,  CDB 
=  25°,  and  CD  =  100  ft. 
Draw  CE  perpendicular 
to  BD.  Then  in  the  right 
triangle  ODE, 


FIG.  10. 


CE=  CD  sin  25°, 
=  100  sin  25°. 


In  the  right  triangle  BCE,  the  angle  CBE  =  10°. 
CE        100  sin  25° 


Hence 


In  the  right  triangle 


sin  10C 


sin  10° 


100  sin  25°  sin  35° 
sin  10° 


CH.  II,  §  9]  RIGHT   TRIANGLES  23 

Applying  logarithms,      log  100  =    2.00000 

log  sin   25°=    9.62595 

log  sin   35°  =    9.75859 

11.38454 

log  sin   10°=    9.23967 
log  AB=    2.14487 
AB  =  139.6. 

SECOND  SOLUTION. — -The  following  is  another  method  of 
solution  in  which  natural  functions  are  used. 

In  the  right  triangle  ABC.  -  =  tan  35°. 

y 

In  the  right  triangle  ABD,  -^ —  =  tan  25°. 

y  + 100 

Solving  this  pair  of  equations  for  aj,  we  have 

x  =  100  tan  35°  tan  25°  =  100  x  .70021  x  .46631 

"  tan  35° -tan  25°  .2339 

Applying  logarithms,         log  100  =    2.00000 

log  .70021=    9.84522 

log  .46631=    9.66868 

11.51390 

log  .2339  =    9.36903 
log  AB=    2.14487 
AB  =  139.6. 

18.  The  shadow  of  a  tower  standing  on  a  level  plane  is  found 
to  be  60  ft.  longer  when  the  sun's  altitude  is  30°  than  when  it 
is  45°.     Prove  that  the  height  of  the  tower  is  30(1  +  V3). 

NOTE.  —  Use  second  method. 

19.  Find  the  height  of  a  chimney  if  the  angle  of  elevation 
of  its  top  changes  from  31°  to  40°  on  walking  toward  it  80  ft. 
'in  a  horizontal  line  through  its  base. 

20.  From  the  top  of  a  cliff  100  ft.  high  the  angles  of  de- 
pression of  two  buoys,  which  are  in  the  same  vertical  plane  as 
the  observer,  are  found  to  be  5°  and  15°.     Find  the  distance 
between  the  buoys. 


CHAPTER   III 
FUNCTIONS  OF  ANY  ANGLE 

10.  Directed  lines.  —  If  a  point  moves  from  A  to  B  in 
a  straight  line,  we  shall  say  that  it  generates  the  line 
AB ;  if  it  moves  from  B  to  A,  it  generates  the  line  BA. 
The  position  from  which  the  generating  point  starts  is 
called  the  initial  point  of  the  line ;  the  point  where  it 
stops,  the  terminal  point.  In  our  study  of  Geometry, 
AB  and  BA  meant  the  same  thing, — the  line  joining  A 
and  B  without  regard  to  its  direction.  But  we  shall 
now  find  it  convenient  to  distinguish  between  AB  and  BA 
by  calling  one  of  them  positive  and  the  other  negative. 

When   either    direction 
along   a   line   has  been 

O ^ >X     chosen  as   the    positive 

FlG  1L  direction    (as     OX    in 

Fig.  11),  then  all  dis- 
tances measured  along  this  line,  or  any  line  parallel  to 
it,  in  this  direction,  shall  be  represented  by  positive 
numbers,  and  those  in  the  opposite  direction  by  negative 
numbers. 

In  Fig.  11,  if  OX  is  chosen  as  the  positive  direction, 
AB  and  MN  are  positive  lines,  while  CB  is  a  negative 
line.  The  measures  of  AB  and  MN  must,  therefore,  be 
positive  numbers ;  but  CB  must  be  represented  by  a 

negative  number. 

24 


CH.  Ill,  §  11]  FUNCTIONS  OF  ANY  ANGLE  25 

The  lines  which  we  shall  use  in  our  further  study  will 
all  be  directed  lines,  unless  the  opposite  is  expressly 
stated,  and  we  shall  be  concerned  not  so  much  with 
the  lines  themselves  as  with  the  measure  of  those  lines. 
We  shall  therefore  find  it  convenient  to  use  the  symbol 
AB  to  represent  "  the  measure  of  the  line  AB  "  (its 
absolute  magnitude  with  its  proper  sign  attached)  ; 
while  if  we  wish  to  speak  of  the  line  itself,  we  shall 
write  "the  line  AB." 

Since  the  lines  AB  and  BA  are  equal  in  magnitude 
but  opposite  in  direction, 

AB  =  -  BA. 

11.  THEOREM.  If  A,  B,  and  O  are  any  three  points  on 
a  straight  line,  ^^  +  BC  =  AC. 


When  the  three  points  are  situated  as  in  Fig.  11,  the 
theorem  is  evident,  since  all  the  numbers  are  positive, 
and  the  measure  of  AC  _  ^ 

equals  the  sum   of   the  A.  _         O"  JS 

measures  of  AB  and  BC. 

In   Fig.    12,   for   the   same   reason   as   above 

AB  =  AC+CB. 

But,  since  we  are  dealing  with  numbers,  we  may 
treat  this  as  an  ordinary  equation.  Hence 

AC=AB-CB 


A 
FIG.  13. 


„. 

" 


In  Fig.  13,  CA  +  AB  =  OB. 

Hence     -AC  +  AB=  -  BC,  or  AC  =  AB  +  BC. 


26  PLANE   TRIGONOMETRY  [Cii.  Ill,  §  12 

Let  the  student  place  the  points  in  other  positions, 
and  show  that  the  theorem  holds  in  all  cases. 

This  theorem  may  be  readily  extended  into  the  fol- 
lowing : 

If  A,  B,  0  •  •  •  J,  K  are  any  number  of  points  on  a  line, 
+  BC+  ...IJ+  JK  =  AK. 


12.  Angles.  —  In  the  previous  chapters  only  angles  less 
than  90°  have  been  considered  ;  but  the  student  is  already 
familiar,  in  his  study  of  plane  geometry,  with  angles  be- 
tween 90°  and  180°.  In  the  further  study  of  mathematics 

it  is  convenient  to  regard  an  angle 
as  formed  by  a  straight  line  revolv- 
ing in  the  plane  about  some  point 
,        t  /  ,       in  the  line.     If  a  line  starts  from 

\u       \s  the  position  OA  and  revolves  in  a 

*        \  / 

\        \          y        /        fixed  plane  about  the  point  0  into 
V4          "^  —  "  /  .    .       . 

the  position  OB,  it  is  said  to  gener- 

ate the  angle  A  OB.     The  position 
from  which  the  moving  line  starts 

is  called  the  initial  side  of  the  angle  ;  the  position  where  it 
stops,  the  terminal  side.  We  shall  regard  the  amount  of 
such  rotation  as  the  measure  of  the  angle.  In  this  sense 
there  is  no  restriction  on  the  size  of  an  angle,  since  there 
is  no  limit  to  the  possible  amount  of  rotation  of  the  mov- 
ing line  ;  after  performing  a  complete  revolution  in  either 
direction,  it  may  continue  to  rotate  as  many  times  as 
we  please,  generating  angles  of  any  magnitude  in  either 
direction. 

Since  there  are  two  directions  in  which  the  moving  line 
may  be  made  to  rotate,  it  is  found  convenient  to  distinguish 


On.  Ill,  §  13]  FUNCTIONS   OF   ANY  ANGLE  27 

between  them  by  using  positive  and  negative  signs,  just 
as  in  algebra  it  is  found  convenient  to  attach  signs  to  the 
ordinary  arithmetic  number.  There  is  no  special  reason 
for  choosing  either  direction  rather  than  the  other  as 
positive ;  but  the  usual  convention  is  to  regard  as  positive 
an  angle  formed  by  a  line  revolving  in  the  direction  op- 
posite to  the  direction  of  rotation  of  the  hands  of  a  clock ; 
the  clockwise  direction  of  rotation  is  then  negative.  In 
reading  an  angle  in  the  ordinary  way  a  letter  on  the 
initial  line  is  read  first ;  for  example,  A  OB  means  the 
angle  formed  by  a  line  in  rotating  from  OA  to  OB,  while 
BOA  means  the  angle  formed  by  a  line  in  rotating  from 
OB  to  OA.  This  method  of  reading  an  angle  is  evidently 
ambiguous,  since  there  is  an  indefinite  number  of  positive 
and  negative  angles,  all  of  which  must  be  read  A  OB.  For 
the  rotating  line,  starting  from  OA,  may  make  any  number 
of  complete  revolutions  in  either  the  positive  or  negative 
direction  and  then  continue  to  the  position  OB,  and  any 
one  of  these  angles  must  still  be  read  A  OB.  Such  angles 
which  have  the  same  initial  and  terminal  sides  are  called 
congruent  angles.  It  will  be  seen  later  that  it  is  usually 
unnecessary  to  distinguish  between  congruent  angles,  since 
their  trigonometric  functions  will  be  found  to  be  the 
same;  but  we  shall  understand  that  the  smallest  of  the 
congruent  angles  is  meant  unless  another  angle  is  indi- 
cated by  an  arrow  in  the  figure. 

13.  The  measure  of  angles.  —  The  angular  unit  of  meas- 
ure with  which  the  student  is  familiar  is  the  degree,  or 
J^th  part  of  one  right  angle.  When  we  wish  to  measure 
an  angle,  we  say  that  it  contains  a  certain  number  of  these 


28  PLANE   TRIGONOMETRY  [Cn.  Ill,  §  13 

units.  This  system  is  convenient  for  numerical  problems 
in  the  solution  of  triangles,  etc.  But  there  is  another 
unit  which  is  almost  universally  used  in  higher  mathe- 
matics. On  the  circumference  of  any  circle  lay  off  an 
arc  AB  equal  in  length  to  the  radius,  and  join  its  ex- 
tremities to  the  centre  of  the  circle.  Since  the  ratio  of 
a  circumference  to  its  radius  is  constant  (and  equal  to 
2  TT),  the  arc  AB  is  always  the  same  fractional  part  of 
a  complete  circumference,  namely 
Then,  since  angles  at  the  cen- 
tre of  a  circle  are  proportional  to 
the  arcs  which  they  subtend,  the 
angle  AOB  is  the  same  fractional 
part  of  four  right  angles,  and  is, 
therefore,  constant. 
FlG>  15'  This  angle  A  OB  is  the  unit  angle 
of  circular  measure,  and  is  sometimes  called  a  radian.  The 
circular  measure  of  any  angle  is  the  ratio  of  that  angle  to 
this  unit  angle,  or  the  number  of  times  the  given  angle  con- 
tains the  unit  angle. 

It  is  usually  written  without  the  name  of  the  unit. 
Since  a  complete  revolution,  or  four  right  angles,  has 
been  shown  to  contain  2  TT  radians,  the  circular  measure 
of  a  right  angle  is  ^;  of  an  angle  of  60°,  ^;  of  45°,  ^.  etc. 

If  a  is  the  circular  measure  of  any  angle  A  OM  and  r  is 
the  radius  of  the  circle, 

AOM     arc^Of     krcJLBf 


_ 

~  r 


Hence  the  circular  measure  of  any  angle  is  equal  to  the 
arc  it  subtends  divided  by  the  radius,  and,  conversely,  the 


CH.  Ill,  §  14]  FUNCTIONS  OF  ANY  ANGLE  29 

length  of  any  arc  equals  the  radius  multiplied  by  the  number 
expressing  the  circular  measure  of  the  angle.  Or  arc  AM=  ar. 
Since  there  are  2  TT  radians  in  a  complete  revolution,  or  360°, 

or»no 

one  radian  —  ——=  57°  17'  45",  approximately. 

2  7T 

We  have  seen  that  it  is  convenient  to  call  angles  formed 
in  the  counterclockwise  direction  of  rotation-  positive  ; 
the  circular  measures  of  such  angles  are,  therefore,  ex- 
pressed by  positive  numbers,  ^,  -^,  etc.  ;  while  the  cir- 
cular measures  of  angles  formed  in  the  clockwise  direction 
of  rotation  are  negative  numbers,  —  —,  —  ^—  ,  etc. 

EXERCISE  IX 

1  .  Express  in  circular  measure  the  angles  15°,  30°,  40°,  120°, 
250°,  300°. 

2.    Express  in  degrees,  minutes,  and  seconds  the  angles  -, 

** 


OK 

7T       Z  7T       O  7T 


7T  7T  7T        K 

9'  T'  ~6~' 

3.  In  a  circle  whose  radius  is  10  inches,  what  is  the  length 
of  an  arc  which  subtends  at  the  centre  of  the  circle  an  angle  of 

3?T       7T       3  7T  r, 

T'  6'  IT' 

4.  In.  a  circle  whose  radius  is  5  inches,  what  is  the  circular 
measure  of  an  angle  at  the  centre  which  subtends  an  arc  of 
10  inches  ? 

14.  Addition  of  directed  angles.  —  If  the  moving  line 
starts  from  OA  (in  any  one  of  these  figures)  and  rotates 
first  through  the  angle  A  OB,  and  then  through  the 
angle  BOO,  it  is  evident  that  the  position  00  which 
the  line  finally  reaches  is  the  same  as  if,  starting  from 


30  PLANE  TRIGONOMETRY  [Cn.  Ill,  §  15 

OA,  it  had  rotated  through  the  single  angle  AGO.  The 
angle  A  00  is  called  the  sum  of  the  angles  A  OB  and  BOO. 
That  is  Z.AOB  +  Z.BOO  =  Z.AOQ. 


By  a  process  similar  to  that  used  in  Art.  11,  it  may  be 
shown  that  the  measure  of  the  sum  of  any  number  of 
angles  is  equal  to  the  sum  of  their  measures,  or  that  the 
above  equation  holds  when,  instead  of  the  angle,  we 
mean,  in  each  case,  the  measure  of  the  angle.  It  will 
not  be  necessary  to  distinguish  further  between  an  angle 
and  its  measure. 

15.  Angles  between  directed  lines.  —  We  shall  have 
occasion  to  speak  of  the  angle  from  the  positive  direc- 
tion of  one  line  to  the  positive  direction  of  another  line, 
and  it  is  convenient  to  have  a  symbol  to  represent  it, 
since  the  ordinary  method  of  reading  an  angle  is  not  suffi- 
cient. We  shall  adopt  the  following  notation  :  {AB,  MN) 
shall  indicate  the  angle  from  the  positive  direction  of 
AB  to  the  positive  direction  of  MN.  (Sometimes  it  is 
more  convenient  to  use  single  letters  to  represent  the 
lines,  as  a  and  6,  when  the  symbol  (a,  6)  will  be  used 
for  the  same  purpose.)  In  this  symbol  it  is  entirely 
immaterial  whether  we  write  AB  or  BA,  MN  or  NM ; 
since  it  is  always  to  indicate  the  angle  between  their  positive 
directions  without  reference  to  the  way  they  are  read. 


CH.  Ill,  §  16]  FUNCTIONS  OF  ANY  ANGLE 


31 


Here  again,  as  in  the 
ordinary  method  of  read- 
ing an  angle,  the  symbol 
is  ambiguous,  since  it  may 
represent  any  one  of  the 
congruent  angles  ;  but 
the  smallest  of  these  will 
be  understood  unless  an- 
other is  indicated  in  the 
figure.  If  the  arrows 
indicate  the  positive  di- 
rections of  the  lines,  in  Fig.  17  (a)  (AB,  MN}  =  AOM, 
a  positive  acute  angle  ;  while  in  Fig.  17  (li)  (AB,  MN) 
=  AON,  a  negative  obtuse  angle. 

16.  Functions  of  angles  of  any  magnitude.  —  We  must 
now  define  the  trigonometric  functions  of  an  angle  of  any 
magnitude.  For  this  purpose,  let  the  plane  be  divided 
into  four  quadrants  by  a  pair  of  indefinite  lines  perpen- 
dicular to  each  other,  —  one  horizontal,  X'X,  called  the 
/-axis,  and  the  other  vertical,  YfY,  called  the  X-axis. 
Let  the  positive  direction  of  the  JT-axis  be  from  left  to 
right,  and  let  the  positive  direction  of  the  JT-axis  be  up- 
ward. Then  all  lines  drawn  parallel  to  these  axes  must 
have  the  same  positive  directions ;  that  is,  a  line  drawn  to 
the  right  or  upward  is  positive,  while  lines  drawn  to  the  left 
or  downward  are  negative.  The  positive  direction  of  any 
line  not  parallel  to  one  of  the  axes  will  be  determined  by 
other  conventions. 

The  point  where  the  axes  cross  is  called  the  origin.  The 
quadrants  are  numbered  as  in  the  figure. 


32 


PLANE   TRIGONOMETRY 


fCn.  Ill,  §  16 


'II 


X 


For  the  purpose  of  defining  the  trigonometric  functions 

of  any  angle,  let  the  angle 
be  placed  with  its  vertex 
at  the  origin  and  with  its 
initial  line  coincident  with 
OX,  the  positive  segment 
of  the  Jf-axis.  It  is  called 
an  angle  in  the  first,  sec- 
ond, third,  or  fourth  quad- 
rants, according  to  the 
position  of  its  terminal 


-x 


IV 


Y 

FIG.  18. 


7T 


line;  that  is,  angles  from  0  to  —  are  angles  in  the  first 

2 

quadrant,  ^  to  TT  in  the  second  quadrant,  etc. 

The  positive  direction  of  segments  measured  along  the 
terminal  line  of  an  angle  will  always  be  from  the  origin 
along  that  terminal  line. 

For  example,  if  we  are  considering  the  angle  XOA,  OA 
is  positive  and  OK  is  negative ;  while  if  we  are  consider- 
ing the  angle  XOK,  OK  is  positive  and  OA  is  negative. 

From  any  point  P  on  the  terminal  line  of  the  angle 
XOA  drop  the  perpendicular  MP  to  the  Jf-axis.  Then 
the  trigonometric  func- 
tions  of  the  angle  XOA 
are  defined  as  follows : 


sin  XOA  = 


cos  XOA  = 


tan  XOA  = 


MP 
OP' 
OM. 
OP' 

MP 
OM 


FIG.  19. 


CH.  Ill,  §  17]  FUNCTIONS   OF  ANY  ANGLE 


33 


The  secant,  cosecant,  and  the  cotangent  are  defined 
as  the  reciprocals  of  the  cosine,  sine,  and  tangent. 

In  each  of  these  definitions  the  direction  as  well  as  the 
magnitude  of  the  lines  is  to  be  considered. 

The  point  P  may  be  taken  as  any  point  either  on  the 
positive  segment,  OA,  of  the  terminal  line,  or  on  OK, 
the  negative  segment  of  that  line.  For  the  similarity  of 
the  triangles  shows  that  the  ratios  are  the  same  numeri- 
cally for  all  positions  of  P ';  and  the  signs  of  the  ratios 
are  also  unchanged  by  a  change  of  P  from  the  positive 
segment  OA  to  the  negative  segment  OK,  since  this 
change  simply  reverses  the  signs  of  MP,  OP,  and  OM. 

From  these  definitions  it  appears  that  the  value  of  any 
function  is  the  same  for  all  congruent  angles,  since  we  are 
concerned  only  with  the  positions  of  the  initial  and  ter- 
minal lines.  The  functions  of  Z  XOA  and  Z  XOK,  which 
differ  by  TT,  are  numerically  equal,  but  may  differ  in  sign. 

17.  Algebraic  signs  of  the  functions.  —  These  definitions, 
when  applied  to  angles  less  than  — ,  will  be  seen  to  agree 

with  the  definitions  in  Chap.  I.     In  the  first  quadrant  all 
the  functions  are  positive,  v 

since  OM,  MP,  and  OP 
are  all  positive.  But  in 
the  other  quadrants  the 

signs  of  some  of  the  func-    v>    m       M       \i/  M  M 

tions  will  be  seen  to  be 
negative.  It  will  be  sim- 
plest always  to  place  P 
on  the  positive  segment  FIG.  20. 


X 


34  PLANE   TRIGONOMETRY  [Cn.  Ill,  §  17 

of  the  terminal  line  of  the  angle,  so  that  OP  shall  always 
be  positive,  and  it  will  only  be  necessary  to  consider  the 
signs  of  OM  and  MP. 

In  the  second  quadrant  MP  is  positive  and  OM  is  nega- 
tive. The  sine  and  cosecant  of  angles  in  the  second  quad- 
rant are,  therefore,  positive  and  all  the  other  functions 
negative. 

In  the  third  quadrant  both  MP  and  OM  are  negative, 
so  that  all  the  functions  of  angles  in  the  third  quadrant  are 
negative  except  the  tangent  and  the  cotangent. 

In  the  fourth  quadrant  OM  is  positive  and  MP  negative. 
The  cosine  and  secant  of  angles  in 
the  fourth  quadrant  are,  therefore, 
positive  and  all  the  other  functions 
negative. 

~  X        The  student  may  find  Fig.  21  use- 
ful in  remembering  the  signs  of  the 
.  functions  in  the  different  quadrants. 

FlQ  21  The  function  written  at  the  head  of 

each  arrow  is  positive  in  the  quad- 
rants through  which  the  arrow  passes. 

EXERCISE  X 

'1  Determine  the  signs  of  the  functions  of  the  following 
angles:  100°,  200°,  300°,  400°,  500°,  000°,  700°,  -50°,  -15()c. 
-350° 

%,    In  which  quadrant  must  an  angle  lie,  if 
(a)  its  sine  and  cosine  are  negative, 
(&)  its  cosine  and  tangent  are  negative, 

(c)  its  sine  is  positive  and  its  tangent  is  negative, 

(d)  its  cosine  is  negative  and  its  tangent  is  positive  ? 


CH.  Ill,  §  18]  FUNCTIONS   OF  ANY  ANGLE  35 

3.  In  which  quadrants  may  an  angle  lie,  if  its  cosine  and 
secant  are  negative  ? 

4.  For  what  angles  in  each  quadrant  are  the  absolute  values 
of  the  sine  and  cosine  the  same  ?     For  which  of  these  angles 
are  they  also  alike  in  sign  ? 

5.  Determine  the  limits  for  x  between  which  sin  x  +  cos  x  is 
positive. 

18.  Functions  of  the  quadrantal  angles.  —  The  angles  0, 
— ,  TT,  -— ,  and  2  TT,  which  have  their  terminal  lines  coinci- 
dent with  one  of  the  axes,  are  called  quadrantal  angles. 
It  is  necessary  to  consider  the  definitions  of  the  functions 
of  these  angles  separately  ;  for  in  some  cases  these  defini- 
tions will  be  found  to  have  no  meaning. 

Let  XOP  be  a  small  angle  and  let  it  decrease.     If  the 
length  of  OP  remains  fixed,  MP 
will    decrease    indefinitely,    and 
the  length  of   OM  will  approach 

that  of    OP.      The  sine 


M 


FIG.  22. 


cosine 


(MP\ 
J  evidently 

decrease     indefinitely     and    ap- 
proach zero  as  a  limit ;  while  the 

(  — — )  approaches  unity  as  a  limit.     Then  sin  0  =  0, 

cos  0  =  1,  and  tan  0  =  0.    Also,  since  sec  A  = -,  sec  0  =  1. 

^^ —  cos  ^ 

But  when  we  consider  the  other  reciprocal  functions,  the 
cosecant  and  cotangent,  we  meet  with  a  difficulty,  since 
the  reciprocal  of  zero  does  not  exist.  There  is,  then,  no 
cosecant  or  cotangent  of  a  zero  angle.  For  small  values  of 


36  PLANE   TRIGONOMETRY  [Cii.  Ill,  §  18 

the  angle  XOP,  the  cosecant     -73     and  the  cotangent 


/  OM\ 

f  -  J  are  large,  and  they  continue  to  increase  without 

limit  as  XOP  approaches  zero.  This  fact  is  usually 
represented  by  writing  esc  0  =  oo  ,  and  cot  0  =  oc  .  But 
these  equations  must  be  distinctly  understood  to  be  abbre- 
viations of  the  statement  that,  as  an  angle  approaches  zero, 
its  cotangent  and  cosecant  increase  indefinitely  ;  they 
must  never  be  understood  to  mean  that  the  cotangent  and 
cosecant  of  a  zero  angle  exists. 

It  is  in  this  sense  only  that  the  symbol  oo  will  be  used 

throughout  this  work.     If,  in  the  expression  -,  x  is  made 

x 

to  decrease  indefinitely,  the  value  of  -  will  increase  indefi- 

x 


nitely.     This  may  be  abbreviated  into  - 


=  oo ,  which 


"1 

should  be  read     -  increases  indefinitely,  as  x  approaches 
x 

zero."  This  is  sometimes  abbreviated  still  further  into 
\  =  oo  .  But  this  must  never  be  interpreted  as  an  ordinary 
equation,  in  which  one  member  is  equal  to  the  other.  It 
has  no  meaning  whatever  except  as  an  abbreviation  of  the 
sentence  above. 

When  it  is  necessary  to  express  the  fact  that  the  nega- 
tive values  of  a  variable  increase  numerically  without 
limit,  the  symbol  —  oc  will  be  used. 

When  the  angle  XOP  increases   toward    —  ,    OM  ap- 

proaches  zero   and   MP  approaches   equality  with    OP. 


Then  the  sine  (-775)  an(i  the  cosecant    -          approach 


CH.  Ill,  §  18]  FUNCTIONS   OF  ANY  ANGLE 


37 


unity ;   the  cosine  f  — -  j  and  the   cotangent  ( )  ap- 

\OP/  fMP\          \MP  J 

proach  zero;    while  the  tangent  (-)   and  the  secant 

fOP\ 

f  — —  j  increase  indefinitely.     Then 


^  =  0,   cot^=0,    esc-  =  1,  tan  x 


and  sec  x 


=  00, 


=  00. 


O    M 


FIG.  23. 


FIG  24. 


It  must  be  here  noted  that,  if  x  is  an  angle  in  the  second 
quadrant  (as  XOP  in  Fig.  24),  OM is  negative,  and  hence 
the  tangent  and  secant  are  negative.  If  x  is  now  made  to 

decrease  toward  — ,  the  negative  values  of  these  functions 

2i 

increase  numerically  without  limit.  This  will  be  found 
to  be  true  in  every  case  where  any  trigonometric  function 
becomes  infinite.  When  the  angle  is  made  to  approach 
this  value  from  one  side,  the  function  has  positive  values 
which  increase  indefinitely,  while  if  the  angle  is  made  to 
approach  this  value  from  the  opposite  side,  the  function  has 
negative  values  which  increase  numerically  without  limit. 


7T 


This  fact  is  usually  expressed   by  writing  tan  — =  ±QD, 

2 


38 


PLANE   TRIGONOMETRY 


[Cii.  Ill,  §  19 


But  the  student  must  keep  constantly  in  mind  the  fact 
that  this  is  only  an  abbreviation  for  the  statement  made 
above. 

Let  the  student  show  in  the  same  manner  that  the  func- 
tions of  the  other  quadrarital  angles  are  as  follows  : 

sin  TT  =  0,  cos  TT  =  —  1,  tan  TT  =  0,  sec  TT  =  —  1,  esc  TT  =  ±  GO, 

cot-TT—  ±  oo. 
sin     TT  =  —  1,     cos     TT  =  0,    tan     TT  =  ±  oo,    sec     TT  =  ±  oo, 


19.  Line  values  of  the  functions.  —  The  trigonometric 
functions  have  been  denned  as  the  ratios  of  lines  to  each 
other,  and  are,  therefore,  abstract  numbers.  But,  by  the 
aid  of  a  circle  drawn  about  the  origin  with  unit  radius, 
lines  may  be  found  which  will  represent,  in  magnitude 
and  direction,  the  values  and  signs  of  the  various  func- 
tions. This  method  of  representing  the  functions  will  be 
found  to  be  an  aid  in  remembering  the  changes  in  sign 

and  value  of  the  func- 

Y  tions. 

T 

Construct      a     circle 

with  its  centre  at  the 
origin,  having  as  radius 
a  line  which  we  shall 
use  as  the  unit  of 
length.  Drop  a  per- 
pendicular from  the 
point  P  where  the  ter- 
minal line  of  the  angle 
meets  the  circle.  Then 


CH.  Ill,  §  20]  FUNCTIONS   OF  ANY  ANGLE  39 


smx,  or  -  ,  is  represented  by  MP  both  in  magnitude 

and  sign,  for  the  denominator  OP  is  always  positive  unity. 

In  like  manner,  OM  will  always  represent  cos  x. 

To  represent  tan  x,  it  is  necessary  to  erect  a  perpen- 
dicular at  A,  in  order  that  the  denominator  of  the  ratio 

AT 

shall  be  the  unit  of  measure.     Then  tan  x  =  —  —  ,  and  AT 

OA 

will  represent  tan  x.  If  the  angle  is  in  the  second  or 
third  quadrant,  the  line  which  is  to  represent  the  tangent 
must  still  be  drawn  at  A  to  meet  the  terminal  line  of 
the  angle,  produced  in  the  negative  direction  ;  for,  if  it 
is  drawn  at  A',  the  denominator  of  the  fraction  would 
be  OA,  which  is  equal  to  —  1,  and  cannot,  therefore,  be 
used  as  the  unit  of  length. 

The  other  functions  may  also  be  represented  by  lines, 
but  it  seems  best  to  think  of  them  as  the  reciprocals  of 
the  three  given  above.  Thus,  the  secant  of  any  angle  has 
the  same  sign  as  the  cosine,  and  in  magnitude  is  the 
reciprocal  of  the  cosine. 

20.  Variations  of  the  functions.  —  The  changes  in  value 
of  the  functions  were  partially  considered  in  Art.  18. 
But  the  student  will  find  it  much  easier  to  remember 
the  changes  in  both  sign  and  value  by  making  use  of 
the  line  values  discussed  in  the  previous  section. 

The  sine.  —  The  line  MP  (Fig.  26)  which  represents  the 
sine  of  XOP  is  seen,  as  the  angle  increases,  to  increase  from 
0  to  1  in  the  first  quadrant  ;  to  decrease  from  1  to  0  in  the 
second  quadrant  ;  to  continue  to  decrease  from  0  to  —  1 
in  the  third  quadrant  ;  and  to  increase  from  —  1  to  0 


40 


PLANE   TRIGONOMETRY 


[Cn.  Ill,  §  20 


P, 


in  the  fourth  quadrant.  This  variation  in  value  and 
sign  may  be  represented  by  the  aid  of  a  curve  as  follows: 

Let  y  =  sin  #,  where 
the  angle  x  is  expressed 
in  circular  measure. 
Then  a  graph  of  this 
equation  may  be  formed 
just  as  is  done  in  the 
case  of  an  ordinary  alge- 
braic equation.  Using 
a  pair  of  perpendicular 
axes,  lay  off  from  0 
along  OX  various  val- 
ues of  #,  and  at  the 
points  thus  determined 

erect  perpendiculars,  whose  lengths  are  the  corresponding 
values  of  y.  For  example,  take  any  convenient  distance,  as 

OA,  to  represent  an  angle  of  — ,  and  at  A  erect  a  perpen- 

2 

dicular  of  unit  length  to  represent  the  fact  that  sin  —  =  1. 
At  B,  midway  between  0  and  A,  erect  a  perpendicular  equal 
to  -|-V2,  or  sin  ,  etc.  Do  this  for  various  values  of  x 


between  0  and  2  TT,  and  then  pass  a  smooth  curve  through 
their  extremities.     This  curve  will  form  a  picture  of  all 


CH.  Ill,  §  20]  FUNCTIONS   OF   ANY  ANGLE 


41 


the  changes  in  value  and  sign  as  the  angle  changes  from 
0  to  2  TT.  It  will  be  seen  that  the  curve  crosses  the  axis 
at  x  =  TT,  since  sin  TT  =  0  ;  that  it  remains  below  the  axis 
from  x  =  TT  to  x  =  2  ?r,  forming  a  curve  like  that  formed 
above  the  axis  from  x=  0  to  #  =  TT;  and  that  this  curve 
will  be  repeated  indefinitely  both  to  the  right  and  left, 
if  x  is  allowed  to  take  on  all  possible  values. 

The  horizontal  distance  which  represents  a  radian,  and 
the  vertical  distance  which  represents  unity  are  usually 
chosen  equal. 

The  cosine.  —  The  l^ine  OM  (Fig.  26),  which  represents 
the  cosine  of  XOP,  is  seen,  as  the  angle  increases,  to 
decrease  from  1  to  0  in  the  first  quadrant  ;  to  decrease 


FIG.  28. 

from  0  to  —1  in  the  second  quadrant;  to  increase  from 
—  1  to  0  in  the  third  quadrant;  and  to  continue  to 
increase  from  0  t/  1  in  the  fourth  quadrant.  Let  the 
student  show  that  this  variation  is  represented  by  the 
figure  given  above.  The  form  of  this  curve  is  seen  to 
be  the  same  as  that  of  the  sine  curve,  but  it  is  moved 

along  the  axis  a  distance  ^  to  the  left. 

2 

The  tangent.  —  It  has  been  shown  that  the  tangent  is 
always  represented  by  a  tangent  to  the  unit  circle  drawn 
from  A.  to  meet  the  moving  radius  produced.  Using  the 


42 


PLANE   TRIGONOMETRY 


[Cn.  Ill,  §  20 


symbol  ±00  in  the  sense  explained  in  Art.  18,  it  is  easily 
seen  that,  in  the  first  quadrant,  AT±  increases  from  0  to 

+  oo  ;  in  the  second  quad- 
rant, AT2  increases  from 
—  oo  to  0  ;  in  the  third 

rp 

quadrant,  AT3  increases 
•X  from  0  to  +00  ;  and  in 
the  fourth  quadrant,  AT± 
increases  from  —  oo  to  0. 
If  we  let  y  =  tan  x,  the 
variation  of  y  will  be 
represented  by  a  curve 
starting  at  0  (Fig.  30)  and  going  upward  indefinitely,  as  x 
approaches  — .  Erect  at  the  point  x  =  —  a  line  MN 


Q  TT  TT  TT 


X 


FIG.  30. 


perpendicular  to  the  axis.     For  every  value  of  x  less  than 


7T 


-  there  are  finite  values  of  the  tangent,  growing  indefi- 


OH.  Ill,  §  20]  FUNCTIONS  OF  ANY  ANGLE  43 

nitely  large  as  x  approaches  — .      But  when  x  =  —,  there 

'—  2 

is  no  tangent.  The  curve,  therefore,  goes  upward  indefi* 
nitely  at  the  left  of  MN,  never  touching  or  crossing  this 
line,  but  constantly  approaching  it.  Such  a  line  is  called 
an  asymptote  of  the  curve.  When  x  is  slightly  greater 
than  ^,  the  tangent  has  a  large  negative  value;  it  in- 
creases toward  0  as  x  approaches  TT  ;  and  increases  indefi- 
nitely as  x  approaches  |TT.  This  branch  of  the  curve  has, 
then,  the  two  lines  MN  and  US  as  asymptotes,  and  crosses 
the  axis  at  the  point  X  =  TT.  When  x  increases  from  |-7r, 
the  tangent  again  starts  with  a  large  negative  value  and 
passes  through  the  same  changes  as  before.  If  x  is 
allowed  to  take  all  positive  and  negative  values,  there 
will  be  a  series  of  such  branches,  just  alike,  at  intervals 
of  TT  from  each  other. 

The  cotangent.  —  Since  the  cotangent  is  the  reciprocal 
of  the  tangent,  it  decreases  from  co  to  0  in  the  first 
quadrant;  decreases  from  0  to  —GO  in  the  second  quad- 
rant; decreases  from  -f- GO  to  0  in  the  third  quadrant; 
and  decreases  from  0  to  —  oo  in  the  fourth  quadrant. 

The  secant.  —  Since  the  secant  is  the  reciprocal  of  the 
cosine,  it  increases  from  1  to  +  GO  in  the  first  quadrant; 
increases  from  —GO  to  —  1  in  the  second  quadrant;  de- 
creases from  —1  to  —GO  in  the  third  quadrant;  and 
decreases  from  +  GO  to  1  in  the  fourth  quadrant. 

The  cosecant.  —  Since  the  cosecant  is  the  reciprocal  of 
the  sine,  it  decreases  from  +00  to  1  in  the  first  quadrant; 
increases  from  1  to  -j-oo  in  the  second  quadrant;  increases 


44  PLANE   TRIGONOMETRY  [Cn.  Ill,  §  20 

from  —  GO  to  —  1  in  the  third  quadrant ;    and  decreases 
from  —1  to  -co  in  the  fourth  quadrant. 


EXERCISE  XI 

1.  How  many  angles  less  than '360°  have  their  cosine  equal 
to  —  |  ?     In  which  quadrants  do  they  lie  ••' 

2.  How  many  angles  less  than  720°  have  their  tangents  equal 
to  5  ?     In  which  quadrants  do  they  lie  '! 

3.  Are  there  two  angles  less  than  180°  which  have  the  same 
sine?   the  same  cosine?   the  same  tangent? 

4.  Construct  the  curve  which  represents  the  change  in  value 
and  sign  of  the  cotangent. 

5.  Construct  the  curve  which  represents  the  change  in  value 
and  sign  of  the  secant ;  the  cosecant. 


CHAPTER   IV 
RELATIONS   BETWEEN  THE  FUNCTIONS 

21.   Relations  between  the  functions  of  any  angle.  —  The 

relations  between  the  functions  of  an  acute  angle  which 
were    proved    in    Art.    5    may  v 

easily  be  shown  to  hold  for 
all  angles.  In  the  triangle 
MOP,  it  is  always  true  that  x- 


M  0 

Y 

FIG.  31. 

By    dividing    successively    by    OP2,    OM2,    and    MP2, 

we  obtain,  as  in  Art.   5, 

sin2  a  +  cos2  a  =  1, 
1  -f-  tan2  «  =  sec2  a, 
1  -f  cot2  a  =  esc2  a. 


It  is  also  apparent  from  their  definitions  that 


MP 

MP<      OP      sin  a  cos  a 

tan  a  =  -——  =  — — :  = ,  and  that  cot «  =  -: 

OM       OM      cos  a  sin  a 


OP 


45 


46  PLANE  TRIGONOMETRY  [Cn.  IV,  §  22 


EXERCISE  XII 

1.  If  sin#  =  —  i,  find  the  possible  values  of  the  other  func- 
tions of  x. 

2.  If  cos  x  =  i,  and  sin  x  is  negative,  find  the  other  functions 
of  x. 

3.  If  tan  x  =  —  3,  and  x  is  an  angle  in  the  fourth  quadrant, 
Hi  id  the  other  functions  of  x. 

4.  If  sec  x  =  4,  find  the  possible  values  of  the  other  functions 
>>£  x. 

5.  If  secA  =  —         — ,  find  the  possible  values  of  the  other 
functions  of  A. 

,^•2    nj2 

6.  If  sin^L  equal  -^ — ^,  find  the  values  of  cos  A  and  cot  A 

x  -\-y 

7.  If  sin 


, 
mr  -f-  2  mn  -+-  2  ?i2 


22.    Functions  of  —  a,  —  ±  a,  IT  ±  a,  -IT  ±  a.  —  In  this 

2  2 

article  we  shall   determine  the  values  of   the   functions 

of  —a,  —  ±  «,  etc.,  in  terms  of  functions  of  a.    In  proving 
2 

these  relations,  it  is  necessary  to  consider  four  cases  ac- 
cording to  the  quadrant  in  which  ex,  lies  ;  but  the  student 
should  first  go  over  the  demonstration  in  the  simplest 
case  where  a  is  an  angle  in  the  first  quadrant,  and  then 
assure  himself  that  the  demonstration  applies  to  all  values 
of  «. 

The  formulas  will  be  obtained  for  the  sine,  cosine,  and 
tangent  only  ;  the  three  reciprocal  functions  are  seldom 
used  and,  if  needed,  formulas  for  them  may  be  obtained 
easily  from  those  given. 


CH.  IV,  §  22]     RELATIONS   BETWEEN  FUNCTIONS 
(a}   Functions  of  —  <*. 


47 


Y                * 

p 

\    1 

A'' 

s\ 

x  x' 

\ 

"NV 

0 

c    | 

M 

M              / 

0   f 

X 

s~ 

N 

P'         > 
P 

/ 

Y' 

X      X 


(a) 


(6)          FIG.  32.          (c) 


In  any  of  these  figures,  let  XOP  =«,  and  XOP1  =  —  a. 
Drop  a  perpendicular  from  P  on  the  Jf-axis  and  continue 
it  to  meet  OPf.  In  either  figure,  the  right  triangles  MOP 
and  MOP1  are  equal,  since  OM  is  common  and  the  geo- 
metrical angles  MOP  and  MOP1  are  equal.  In  every  case 
OM  is  identical  for  the  two  angles,  OP1  =  OP,  and  MP' 
=  -MP. 

MPr      —  MP 
Hence    sin  (-  a)  =  -^^^  =     ^    =  -  sin  a, 


OP1 


cos  (—  «)  = 


tan  (—  «)  = 


OP' 


OP 

OM 
OP 


OM         OM 


=      cos  a, 


=  —  tan  «. 


Here  XOP  may  be  the  angle  indicated  by  the  arrow, 
or  it  may  be  any  one  of  the  positive  or  negative  congruent 
angles  which  are  read  in  the  same  way.  We  must  then 
understand  by  XOP'  that  one  of  the  congruent  angles 
which  is  equal  to  —  XOP.  For  example,  XOP  may  be 
the  negative  angle  formed  by  revolving  from  OX  to  OP 
in  the  negative  direction.  Then  XOP'  is  the  positive 
angle  formed  by  revolving  from  OX  to  OP'. 


48 


PLANE   TRIGONOMETRY 


[Cn.  IV,  §  22 


The  above  demonstration  is  therefore  general,  and  a 
may  be  any  angle,  positive  or  negative.  These  state- 
ments apply  also  to  the  demonstrations  which  follow. 


Functions  of  ^  ±  «. 

jL 


(6)        FIG.  33. 


(c) 


(d) 


7T 


In  any  of  these  figures,  let  XOP  =  a  and  XOPf  =  -|  +«. 


If  OP'  is  taken  equal  to  OP,  the  triangles  .ftfOP  and 
M'OP1  are  equal  ;  for  the  geometrical  angles  MOP  and 
M'P'O  are  equal,  having  their  sides  perpendicular. 
Then  the  sides  of  these  triangles  are  equal  in  magnitude, 
but,  when  their  signs  are  considered,  OM'  =  —  MP,  and 
M'P'  =  OM. 

OM 


7T     ,  M'Pf 

Hence  sm[-+«    = 


OP 


=  cos 


-  MP 


=  —  cot  a. 


Since  the  demonstration  just  given  holds  for  all  values 
of  ce,  we  may  replace  a  by  —  a. 


CH.  IV,  §  22]     RELATIONS   BETWEEN   FUNCTIONS 


49 


7T 


Hence        sin  [  —  •—«)  =s       cos  (  —  a)  =  cos  a, 


cos 


(  ^  —  CM  =  —  sin  (  —  a)  =  sin  a, 


tan  (  -  —  a  )  =  —  cot  (  —  a)  =  cot  a. 
\2        / 


Functions  of  TT  ±  a. 


0          M 


Y 
(O) 


M 


P 

FIG.  34. 


Y' 


<<9 


111  either  of  the  given  figures  let  XOP  =  a,  and 
'  =  7r  +  a.  If  OP'  is  taken  equal  to  OP,  the  tri- 
angles If  OP  and  Jf'OP'  are  equal,  and  M'P'  =  -MP, 
and  OM'=-OM. 

M'  P'      —  MP 
Hence  sin  (TT  +  a)  =  -^^  =     ^     =  -  sin  .a, 


OP' 


OP 


,       Olf '       -  OM 

cos  (TT  -f  a)  = =  — ^7—  =  —  cos 


OP' 


Olf' 


OP 


OM 


=      tan  a. 


Replacing  a  by  —  a,  we  have 

sin  (TT  —  «)  =  —  sin  (—  a)  =  sin  a, 
cos  (TT  —  a)  =  —  cos  (—«)=  —  cos  a, 
tan  (TT  —  a)  =  tan  (—«)=  —  tan  a. 


37" 


50  PLANE   TRIGONOMETRY  [Cn.  IV,  §  23 

(d)  Functions  of  |  TT  ±  a. 

Let  the  student  construct  figures  for  the  other  values  of 
a  and  show  that 

sin  (|  TT  +  a)  =  —  cos  os, 

* 

cos  (f  TT  +  «)  =      sin  oc, 

tan  (|-  TT  -}-  «)  =  —  cot  a 

Also  that 

sin  (|  TT  —  «)  =  —  cos  «, 
cos  (|^  TT  —  «)  =  —  sin  a, 
tan  (|  TT  —  «)  =  cot  a. 

(e)  Functions  of  2  TT  ±  «. 

The  functions  of  2  TT  —  cc  are  equal  to  the  same  functions 
of  —  #,  since  these  angles  are  congruent.  For  the  same 
reason  the  functions  of  2  HTT  +  a  (where  n  is  any  integer 
and  a  is  any  angle,  positive  or  negative)  are  equal  to  the 
same  functions  of  «. 

23.  Reduction  of  the  functions  of  any  angle  to  functions 
of  an  angle  less  than  — .  —  It  appears  from  the  results  of 

the  previous  article  that  the  functions  of  any  angle  may 
be  obtained  in  terms  of  the  functions  of  an  angle  less 
than,  or  equal  to,  45°.  This  may  be  done  by  the  aid  of 
the  formulas  there  derived  ;  but  since  the  functions  of 
angles  in  the  first  quadrant  are  all  positive,  it  is  best 
to  use  the  following  simple  rule,  which  is  easily  derived 
from  the  formulas  of  the  previous  article  : 

If  180°  or  360°  is  subtracted  from  a  given  angle,  or  if  the 
given  angle  is  subtracted  from  180°  or  360°  (so  as  to  obtain 


CH.  IV,  §  23]     RELATIONS  BETWEEN  FUNCTIONS  51 

in  either  case  an  acute  angle),  the  functions  of  the  resulting 
angle  will  be  numerically  equal  to  the  same  named  functions 
of  the  given  angle  ;  while  if  the  given  angle  is  combined  in 
the  same  way  ivith  90°  or  270°,  the  functions  of  the  resulting 
angle  will  be  numerically  equal  to  the  co-named  functions  of 
the  given  angle. 

In  any  case,  attach  to  the  result  the  proper  sign  of  the 
function  of  the  given  angle,  according  to  the  quadrant  in 
which  it  lies. 

'  For  example,  to  obtain  sin  290°.  This  is  equal  numeri- 
cally to  cos  20°  ;  but  since  290°  is  an  angle  in  the  fourth 
quadrant  and  the  sine  is  negative  in  that  quadrant, 
sin  290°  =  -  cos  20°. 

Any  multiple  of  360°  may  be  added  to,  or  subtracted 
from,  an  angle  without  changing  the  value  or  sign  of  any 
of  its  functions. 

EXERCISE  XIII 

1.  Express  each  of  the  following  functions  in  terms  of  func- 
tions of  angles  not  greater  than  45°. 

(a)  sin  100°,  (e)    sin  -110°,  (i)    cos  395°T~ 

(6)   cos  245°,  (/)  cos  -125°,  (j)  tan  560°, 

(c)   tan  310°,  (g)  tan  -335°,  (&)  sin  ITT, 

sec  190°,  (A)  esc  -25°,  (I)   COS|TT. 


2.    Find  all  the  functions  of  each  of  the  following  angles. 
(See  Art.  4.) 

(a)  120°,  (e)   225°,  (i)      330°, 

(b)  135°,  (/)240°,  (j)  -30°, 

(c)  150°,  (g)  300°,  (fc)   -45°, 

(d)  210°,  (7i)  315°,  (I)    -60°. 


52  PLANE    TRIGONOMETRY  [Cn.  IV,  §  24 

v    3.    Prove  geometrically  the  formulas  for  the  functions  of 

37T 


^  —  a  \,  (TT  —  a),  and  j  — a  J  in  terms  of  functions  of  a. 

4.    From  the  formulas  for  the  functions  of  —a  and  (  --f  a\ 
derive  algebraically  the  formulas  for  the  functions  of  (?r  ±  a) 


and  ^  ±«J. 

SUGGESTION. — sin  (TT  —  a)  =  sin    -  4-  (  -  —  a  )    =  cos  ( -  —  a 

=  sin  a. 

/          \ 

5.  Obtain  the  functions  of  Y  a j  in  terms  of  the  functions 

of  a. 

6.  Obtain  the  functions  of  (a  —  TT)  in  terms  of  the  functions, 
of  a. 

24*.  Projection.  —  The  projection  of  a  point  on  a  line 
is  the  foot  of  the  perpendicular  dropped  from  the  point 
to  the  line. 

The  projection  of  one  line  on  another  is  the  locus  of 
the  projections  of  its  points,  or  the  distance  measured 

along  the  second  line  from  the 
projection  of  the  initial  point 
of  the  first  line  to  the  pro- 
jection of  its  terminal  point. 
-  In  Fig.  36,  CD  is  the  pro- 


c      FIQ  ^      D  jection   of   AB  on    OX.      The 

direction   as  well   as  the  mag- 

nitude of  CD  must  be  considered,  and  CD  will  be 
positive  or  negative  according  as  it  is  drawn  in  the 
positive  or  negative  direction  of  OX. 

Throughout    the    present    work    the    projection    will 
always  be  upon  two  perpendicular  axes,  X'X  and  Y'  Y. 


CH.  IV,  §  25]     RELATIONS  BETWEEN   FUNCTIONS  53 

When  we  wish  to  speak  of  the  projection  of  a  line,  as 
AB,  on  the  JT-axis,  it  will  be  written  projx  AB ;  on  the 
I^-axis,  projy  AB.  These  symbols  will  always  mean  the 
distance  measured  along  the  axes  from  the  projection  of  A 
to  the  projection  of  B,  and  will  be  positive  or  negative 
according  as  they  are  drawn  to  the  right  or  upward, 

or  to  the  left  or  downward. 

V 

25*.  Projection  of  a  broken  line.  —  The  sum  of  the  pro- 
jections on  any  axis  of  any  series  of  lines,  AB,  BC,  CD, 
etc.,  in  which  the  initial  point  of  each  line  is  joined  to  the 
terminal  point  of  the  preceding  line,  is  equal  to  the  projec- 
tion on  the  same  axis  of  AD,  the  line  which  joins  the 
initial  point  of  the  first  line  with  the  terminal  point  of 
the  last  line. 

In   Fig.  37,  ab  =  proj^ AB,    bc=p?ojxBC,    etc.      But 

by  Art  U'  ab  +  be  +  cd  =  ad. 

Hence      pro].,.  AB  +  proj^  BO  +  projx  CD  =  projx  AD. 


If  the  terminal  D  of  any  such  broken  line  coincides 
with  the  initial  point  A,  ad  =  0,  and  we  see  that' 
the  projection  of  any  closed 
contour  is  zero. 

The  lines  AB,  BO,  etc., 
are  directed  lines,  but  it 
is  not  necessary  that  their 
positive  direction  should  be 
from  A  to  B,  etc.  The 
measure  of  some  of  the  lines  as  read  may  be  positive 
and  of  others  negative  without  affecting  the  truth  of 
the  theorem. 


PLANE   TRIGONOMETRY 


[CH.  IV,  §  26 


26*.  Projections  on  the  axes  of  any  line  through  the 
origin. — The  definitions  of  the  functions  may  be  given 
very  concisely  by  the  aid  of  projection.  Let  P  be  any 
point  on  either  the  positive  or  negative  segment  of  the 
terminal  side  of  the  angle  XOA.  The  angle  (OJT,  OP) 
is  then  the  same  as  the  angle  XOA.  Also  OM =  proj^  OP, 
and  MP  =  proj?/  OP.  Substituting  these  expressions  in 

the  definitions  of  the  sine 
and  cosine  given  in  Art.  16, 
they  become 


M 


, 

and 


FIG.  38. 


OP)  = 


OP 


Clearing  of  fractions,  we  have 

proj^OP  -  OP  cos(OJf,  OP), 
and  pro }y  OP  =  OP  sin  (  OX,  OP) . 

The  student  must  remember  that  in  these  formulas 
proj^OP  and  proj^OP  are  the  measures  of  the  distances 
from  the  projection  of  0  to  the  projection  of  P,  the  signs 
being  determined  by  the  directions  of  the  axes ;  that  OP 
is  the  measure  of  the  line  OP ;  and  that  (  OX,  OP)  is  the 
angle  from  the  positive  direction  of  OX  to  the  positive 
direction  of  OP. 

27*.  Projections  of  any  line  on  the  axes. — We  shall 
now  proceed  to  obtain  the  more  general  formulas  for  the 
projection  of  any  line,  AB,  on  the  axes. 

Through  A,  which  may  be  either  the  positive  or  nega- 
tive extremity  of  the  line  AB,  draw  a  pair  of  axes,  AX' 


CH.  IV,  §  27J     RELATIONS  BETWEEN   FUNCTIONS 


and  AY',  parallel  to  the 
given  axes.  The  pro- 
jections of  AB  on  these 
new  axes  are  evidently 
equal  in  magnitude  and 
direction  to  its  projec- 
tions on  the  original 
axes,  OX  and  OY ';  and 
the  angles  made  by  AB 
with  these  axes  are  the  same  as  those  made  by  it  with 
the  original  axes.  Then,  from  the  formulas  obtained  in 
the  previous  article, 

=  AB  cos  (OX,  AB),  [9] 

=  ABsin  (OJT,  AB).  [10] 


FIG.  39. 


EXERCISE  XIV 

1.  What  are  the  projections  on  the  axes  of  a  line  5  inches 
long  which  makes  an  angle  with  the  X-axis  of  30°  ?   of  100°  ? 
of  200°  ? 

2.  If   proja!AB  =  3,  and  projy  AB  =  —  4,  find  the  length  of 
AB  and  the  angle  (OX,  AB). 

3.  If  A  B  =  10,  and  proj^JB  =  -  3,  find  the  angle  (OX,  AB). 

4.  Describe    an    equilateral   triangle    (each   side   being   10 
inches  in  length)  by  going  from  A  to  B,  then  to  C,  and  back 
to  A.    What  is  the  angle  (AB,  BC)  ?    (AB,AC)?    (BC,AB)? 

What  is  the  projection  of  AC  on  AB?   of  BC  on  AB?   of 
CA  on  BC? 

5.  In  the  triangle  of  problem  4  drop  a  perpendicular  CD 
from  C  on  AB.     Let  its  positive  direction  be  from  D  to  C. 
What  is  the  projection  of  BC  on  this  perpendicular  ?   of  CA  ? 
of  AB  ?     What  is  the  projection  of  DC  on  BC?   on  AC? 


56 


PLANE   TRIGONOMETRY 


[CH.  IV,  §  28 


28*.  Functions  of  the  sum  and  difference  of  two  angles.  — 
We  shall  now  proceed  to  find  the  functions  of  the  sum 
and  difference  of  two  angles  in  terms  of  the  functions  of 
those  angles.  Let  the  first  angle  a  be  placed  in  the  posi- 
tion XOA,  and  let  the  second  angle  /3  (=AOB}  be  added 
to  a,  making  «  +  (3  ( =  XOB) .  From  any  point  P  of  the 


FIG.  40. 


terminal  line  OB  of  /3,  drop  a  perpendicular  KP  on  OA. 
The  positive  direction  of  all  lines  in  the  figure  except  KP 
is  fixed  by  our  former  conventions  ;  for  we  have  agreed 
that  the  positive  direction  of  the  initial  and  terminal  lines 
of  an  angle  shall  be  from  the  vertex  along  those  lines. 
To  determine  the  positive  direction  of  KP,  draw  from  0 

a  line   OR,  making  with   OA  a  positive  angle  —  •     Then 

OR  bears  the  same  relation  to  OA  that  OY  bears  to  OX; 
and  for  any  angle  having  OA  as  its  initial  line,  the  posi- 
tive direction  of  KP,  a  perpendicular  to   OA,  must   be 
taken  parallel  to  OR. 
From  Art.  26  we  have 


But  by  Art.  25  the  projection  of   OP  on  the  !F-axis  is 


CH.  IV,  §  28]     RELATIONS   BETWEEN   FUNCTIONS  5,7 

equal  to  the  sum  of  the  projections  of  OK  and  KP  on  the 
same  axis.     Hence 


or  by  [10], 


=  -       [  <9JTsin(  OX,  OK)  +  KP  sin  (OX,  KP)~]  . 


The  angle  (OX,  OK)  is  seen  at  once  to  "be  a,  and  since 
the  positive  direction  of  KP  is  the  same  as  that  of  OR, 

(OX,  KP)  =  XOR  =  «  +  -• 

Hence  sin  (  OX,  KP)  =  cos  «. 

d  K  JfT* 

Noting  that  •—  —  =  cos  /3  and  —  —  =  sin  (3,  we  have 

sin  (a  +  P)  =  sin  a  cos  p  +  cos  a  sin  p.  [11] 

In  like  manner, 


Here  cos  (OX,  OK)  =  cos  «, 

and  cos  (  OX,  KP)  =  cos  fa  +  —  J  =  —  sin  a. 

Hence          cos  (a  +  p)  =  cos  a  cos  p  -  sin  a  sin  p.  [12] 


58 


PLANE   TRIGONOMETRY 


[On.  IV,  §  28 


Since  the  magnitudes  and  directions  of  all  the  lines  and 
angles  in  the  figure  have  been  carefully  considered  in  this 
demonstration,  it  holds  for  all  cases.  But  in  order  that 
the  student  may  assure  himself  of  this  fact,  let  him  draw 
various  figures  according  to  the  same  directions,  using  the 
angles  of  different  magnitudes  (for  example,  see  Fig.  42), 
and  go  through  this  demonstration  carefully,  making  sure 
that  every  statement  made  above  applies  to  every  figure. 


Y     A 


Let  him  also  take  note  that  the  demonstration  applies 
letter  for  letter  when  fi  is  a  negative  angle  (see  Fig.  43). 
We  may,  therefore,  replace  ft  by  —  fi  in  formulas  [11] 
and  [12]  and  obtain  the  following  formulas  for  the  sine 
and  cosine  of  the  difference  of  two  angles  : 


sin  [«  +  (  —  /3)]  =  sin  a  cos  (—  /3)  +  cos  a  sin  (—  /3), 

sin  (a  —  p)  —  sin  a  cos  p  —  cos  a  sin  p, 
cos  [«  +  (—  /3)]  =  cos  a  cos  (—  fi)—  sin  a  sin  (—  /3), 

cos  (a  -  p)  =  cos  a  cos  p  +  sin  a  sin  p.  [14] 


CH.  IV,  §  29]     RELATIONS   BETWEEN   FUNCTIONS 


59 


29 .    Second  method  of  finding  sin  (a  +  p )  and  cos  (a  +  p) .  — 

The  following  method  of  finding  the  formulas  for  sin(«  +  £) 
and  cos  («  +  /8)  avoids  the  use  of  projection  and  may  be  pre- 
ferred by  some  teachers. 

Let  Z  XOA  =  «,  and 
ZAOB=/3.  Then  ZX  OB 
=  a  -h  @.  Through  any 
point  P  of  the  line  OB 
draw  PM  perpendicular 
to  the  X-axis  and  PTTper- 
pendicular  to  the  line  OA. 
Through  K  draw  .BT^V per- 
pendicular to  the  Jf-axis 
and  LK  parallel  to  the  same  axis. 


FIG.  44. 


Prolong  LK  toward 


the  right  to  S.     Then  Z  SKA  =  a  and  Z  SKP  =  a  +  90°. 


Then        sin 


MP  =  NK    LP 
OP       OP      OP' 


NK 
OK 


OK 
OP 


LP      KP 
KP  '   OP 


But 


NK 
OK 


=  sin  XOA  =  sin  ce, 


=  cos/3, 


=  sin  SKP  =  sin(«  +  90°)  =  cos  a, 


OP 


=  sin/3. 


Hence 


sin  (a  +  p)  =  sin  a  cos  p  +  cos  a  sin  p, 


pi] 


60 


PLANE   TRIGONOMETRY 


[Cn.  IV,  §  29 


AI 
Also 


But 


0.       OM     ON  ,  NM 

COB(«  +  0)  =  —  =—  +^ , 

=  ON      OK     KL 

OK'  OP    KP 

ON 

-—-  =  cos  XOA  =  cos  a, 
OK 


=  cos  ACS  =  cos 


KP 
OP' 


~ 


-  cos 


=  cos  (a  -f  90°)  =  -  sin  a, 


KP 
OP 


=  sin  AOB  =  sin 


Hence 


COS  (a  +  P)  =  COS  a  COS  P  -  Sill  a  sin  p. 


In  Fig.  44  not  only  are  a  and  /3  acute,  but  their  sum 

a  +  /3  is  also  acute.     But  the  proof  applies  without  change 

B  to  Fig.  45,  in  which  a  -f  /3  is 

obtuse.  The  only  difficulty 
which  the  student  is  likely 
to  meet  is  the  equation 

OM=ON+NM, 

_  which  is  used  in  finding 
cos  («  +  /3)  ;  but  this  is  seen 
to  hold  in  Fig.  45,  when  the 
direction  as  well  as  the  magnitudes  of  the  lines  is  considered. 
By  a  very  slight  change,  this  proof  may  be  made  per- 
fectly general,  so  that  it  will  apply  to  all  values  of  a  and 
/3.  Through  0  draw  OR,  making  a  positive  right  angle 
with  OA,  and  having  its  positive  direction  from  0  toward 
R.  Let  KP,  which  is  parallel  to  OR,  have  the  same 
positive  direction  as  OR. 

If,  in  the  above  demonstration,  we  replace  Z.SKP  by 


0       N 
FIG.  45. 


CIL  IV,  §  29]     RELATIONS   BETWEEN   FUNCTIONS 


Gl 


T,  KP),  we  shall  have  taken  account  of  the  directions 
of  all  lines  and  angles,  and  the  demonstration  will,  there- 
fore, be  perfectly  general,  holding  for  all  values  of  a  and 
ft,  both  positive  and  negative. 

For  example,  consider  Fig.  46,  in  which  a  and  /3  are 
both  obtuse.  The  construction  is  the  same  as  above. 
From  P,  any  point  of  the 
terminal  line  of  ft,  draw 
PM  perpendicular  to  the 
X-axis  and  PK  perpen- 
dicular to  the  line  OA, 
produced  through  the 
origin .  Through  K  draw 
KN  perpendicular  to  the 
X-axis  and  LK  parallel 

to  the  same  axis,  prolong- 

~r  ^7'         t  i  c*  FIG.  46. 

ing  LK  to  the  right  to  S. 

Draw  OR,  making  a  positive  right  angle  with  OA.  Let 
the  student  now  follow  the  demonstration  on  page  59. 
The  first  statements  are  evident.  In  considering  the 
functions  of  **,  since  K  is  on  the  terminal  line  extended 
through  the  origin,  OK  is  negative,  but  by  definition 

sin  a=  — — ,  and  cos  a= In  determining  the  functions 

OK  OK 

of  ft,  turn  the  book  so  that  OR  points  upward.     Then  it 

KP  OK 

is  seen  that,  by  definition,  sin  ft  = -,  and  cos  ft  =  — —  • 

In  determining  the  functions  of  (SK,  KP),  think  of  K 
as  a  new  origin  and  LS  as  a  new  X-axis.     Then 

KL 
KP 


sin  (tfJT,  KP)  =        ,  and  cos  (SK,  KP)  = 


62  PLANE    TRIGONOMETRY  [Cn.  IV,  §  30 

The  student  should  have  no  difficulty  in  seeing  that  all 
the  other  equations  hold  for  this  figure.  Let  him  also 
construct  other  figures  according  to  the  same  directions 
and  go  through  the  demonstration  carefully,  making  sure 
that  every  statement  made  above  applies  to  every  figure. 

Since  this  demonstration  holds  for  all  values  of  a  and 
/3,  we  may  replace  ft  by  —  ft  and  obtain  the  following 
formulas  for  the  sine  and  cosine  of  the  difference  of  two 
angles  : 

sin  [a  +  (  —  /3)]  =  sin  «  cos  (—/&)  +  cos  a  sin  (  —  y#), 

sin  (a  -  P)  =  sin  a  cos  p  -  cos  a  sin  p,  [13] 


cos  [<*  +  (  —  /3)]  =  cos  acos(—  /3)—  sin  a  sin(—  /?), 

cos  (a  -  p)  =  cos  a  cos  p  +  sin  a  sin  p.  [14] 

30.   tan(a±p). 
Since      ten 


we  have     tanQ  +  cos*  sin 


cos  a  cos  p  —  sin  a  sin  /3 

But  this  may  be  expressed  entirely  in  terms  of  the  tan- 
gent by  dividing  both  numerator  and  denominator  by 
cos  a  cos  /3.  This  gives 


In  like  manner,  let  the  student  show  that 

tan  (a  _p)      tan  a  -tan  P  [1G] 

1  +tanatanp 


CH.  IV,  §  30]     RELATIONS   BETWEEN  FUNCTIONS  63 

EXERCISE  XV 

1.    Find  sin  75°  from  the  functions  of  30°  and  45°. 
SOLUTION.  —  sin  75°  =  sin  (30°  +  45°), 

=  sin  30°  cos  45°  +  cos  30°  sin  45°. 


=  J(V2+V6). 

2.  Find  cos  75°. 

3.  Find  tan  75°. 

4.  Find  sin  15°,  cos  15°,  tan  15°. 

5.  If  sin  a  =  ^  and  sin  ft  =  \,  find  sin  (a  +  ft),  when  a  and  ft 
are  both  acute  :)  when  they  are  both  obtuse. 

6.  By  the  aid  of  formulas  [11]  to  [14],  prove  the  various 

formulas    for    the    functions    of   (^  +  a)»    (TT  ±  «)?  etc.     (See 
Art.  22.) 


7.  Find  sin( 

SOLUTION.  —  Replacing  ft  in  formula  [11]  by  ft  -f-  y,  we  have 
sin  [«  +  (/3  +  y)]  =  sin  «  cos  (ft  -f-  y)  +  cos  a  sin  (ft  +  y) 
=  sin  a  (cos  J3  cos  y  —  sin  ft  sin  y)  + 
cos  a  (sin  /?  cos  y  4-  cos  /?  sin  y) 
=  sin  a  cos  /?  cos  y  +  cos  a  sin  ft  cos  y  •+• 
cos  a  cos  /2  sin  y  —  sin  a  sin  ft  sin  y. 

8.  Find  cos  (a  -f-  0  —  y). 

9.  Find  tan  («  —  0  -f  y). 

Transform  the  first  member  into  the  second. 

10.  sin  (a  4-  ft)  sin  («  —  /?)  =  sin2 «  —  sin2/?. 

11.  cos  (a  +  /?)  cos  (a  —  ft)  =  cos2  a  —  sin2  ft. 


64  PLANE  TRIGONOMETRY  [Cn.  IV,  §  31 

12.  cos  g-«)cosg-0)-sing-«)  sing-/?) 

=  sin  («  -f  0). 

13.  cos  (a  -f  ft)  cos  a  +  sin  a  sin  (a  -f  /?)  =  cos  ft. 

tan  «  —  tan  6  / 

14 .  —  — £  =  tan  («  —  fl)  tan  a. 
cot  «  +  tan  ft 

31.  Functions  of  twice  an  angle.  — If  in  formulas  [11] 
and  [12]  we  place  /8  =  «,  they  become 

sin  («  +  «)=  sin  «  cos  a  +  cos  «  sin  a, 

or  sin  2  a  =  2  sin  a  cos  a,  [17] 

and  cos  (a  -|-  «)  =  cos  a  cos  «  —  sin  a  sin  a, 

or  cos  2  a  =  cos2  a  -  siii2  a.  [18,  a] 

By  replacing  cos2  a  by  1  —  siii2«,  [18,  a]  becomes 

cos2a^l-2sin2a.  [18,  6] 

Also,  by  replacing  sin2  a  by  1  —  cos2  «,  it  becomes 

cos2a  =  2cos2a- 1.  [18,  c] 

These  three  forms  for  cos  2  a  should  be  remembered,  and 
the  one  most  convenient  for  the  purpose  in  hand  should 
be  used. 

If  we  place  ft  =  a  in  formula  [15],  it  becomes 

f  tan  a  +  tan  a 

tan    «  + 


1  —  tan  a  tan  a 

Q  tilH   fL  r--<  f\-^ 

or  tai,2a=    -    V—.  [19] 


(JH.  IV,  §  31]     RELATIONS   BETWEEN  FUNCTIONS  65 

The  functions  of  3  a,  4  a,  etc.,  may  be  found  in  a  similar 
manner  by  letting  ft  =  2  «,  etc. 

These  equations  express  the  relations  between  the  func- 
tions of  any  angle  and  the  functions  of  twice  that  angle. 
It  is  not  necessary  that  the  angle  in  the  first  member 
should  be  expressed  as  2  a  and  that  in  the  second  as  a.  It 
is  only  necessary  that  the  angle  in  the  first  member  should 
be  twice  that  in  the  second.  We  may  then  replace  2  a  by 
a,  if  at  the  same  time  we  replace  a  by  $  «.  These  formulas 
may  then  be  written 

sin  a  =  2  sin  ^  a  cos  ^  a,  [20] 

cos  a  =  cos2  *  a  -  sin2  \  a,  [21,  a] 

=  l-2sin2ia,  [21,  ft] 

=  2cos2|a-l,  [21,  c] 


tana- 


[22] 


EXERCISE  XVI 

1.    Find  sin  2  a,  if  cos  a  =  i-. 

V%.    Find  cos  2  a,  if  tan  a  =  3. 

\s  3.    Find  tan  2  a,  if  sin  a  =  .6. 

4.    Find  sin  3  a. 
SOLUTION, 

sin  3  a  =  sin  (a  -f  2  a)  =  sin  a  cos  2  «  +  cos  a  sin  2  a, 
=  sin  a  (1  —  2  sin2«)  -f  cos  a  •  2  sin  a  cos  a, 
=  sin  a  (1  —  2  sin2«)  -f-  2  sin  «  (1  —  sinV), 
=  3  sin  a  —  4  sin3a. 


66  PLANE   TRIGONOMETRY  [On.  IV,  §  32 

*""  5.  Find  cos  3  a  and  tan  3  «. 

6.  Find  sin  4  a,  cos  4  a,  and  tan  4  a. 

7.  Find  sin  5  «,  and  cos  5  «. 

Transform  the  first  member  into  the  second. 

Is*  8.  2  cot  2  «  =  cot  a  —  tan  rc. 

9.  2  esc  2  a  —  tan  «  +  cot  a. 

/^  10.  esc  2  «  +  cot  2  «  =  cot  «. 


11.    -  )  -  -  =  esc  2 
l-tan2[--a 
V4 

_  _     cos  «  4-  sin  «     cos  «  —  sin  <*      0  ,       0 
i  &  .    -  ;  ---  ;  -  =  £>  tan  &  cc. 
cos  «  —  sin  a      cos  «  +  sin  a 

2  cot  rc 
13.    tan  2  a  = 


cor«  —  1 

14.     2  CQt  2  «  =  (1  +  tan  a)  cot  a. 
1  —  tan  a 

1  ±  sin2« 
cos  2  a 


L-16.    tan  [-  +  «  ]  +  tan  [  -—  aW  2  sec  2 
4  4 


32.    Functions   of   half  an  angle.  —  It  has  been  shown 
in  the  previous  article  that 

cos  a  =  1  —  2  sin2  1  a. 
Solving  this  for  sin  |  a,  we  have 

[23] 


Cn.  IV,  §  32]     KELATIONS  BETWEEN  FUNCTIONS  67 

Also  from  formula  [21,  c], 

cos  a  =  2  cos2  £  a  —  1. 

Solving  for  cos  J  «,  we  have 

i  /I  +  cos  ct  ro  j  -\ 

COS-a  =  ±\-^ •  [/4J 

Dividing  [23]  by  [24],  we  have 

[25,  a] 


-    I 

- 


sin  a 


(1  +  cos  a)2      1  +  cos  a 

These  three  forms  for  tan  |  «  should  all  be  remembered, 
and  the  one  most  convenient  for  the  purpose  in  hand 
should  be  used. 

It  must  be  noted  that  the  sign  ±  before  the  radicals 
does  not  mean  that  it  has  two  values  for  any  given 
angle  a,  but  that  it  is  impossible  to  determine  in  gen- 
eral what  sign  should  be  used.  In  any  particular  case, 
determine  the  quadrant  in  which  |  a  lies,  and  affix  the 
proper  sign  for  each  function.  The  ambiguous  sign  is 
omitted  before  the  last  two  forms,  since  1  —  cos  a  and 
1  -f-  cos «  are  always  positive,  and  tan  |  «  may  easily  be 
shown  always  to  have  the  same  sign  as  sin  a. 

It  is  sometimes  convenient  to  express  the  functions 
of  a  in  terms  of  the  functions  of  2  a.  This  may  be  done 


68  PLANE   TRIGONOMETRY  [Cn.  IV,  §  32 

by  replacing  |  a  by  a,  and  a  by  2  a  in  the  above  formulas. 
They  then  become 

^,  [26] 


/I  +  cos  2  a 

[27] 
[28,  a] 
[28,  6] 
T28.  /-I 

V        2 

.  /I  —  COS  2  a 

*  1  +  COS  2  a' 
1  —  COS  2  a 

sin  2  a    : 
sin  2  a 

EXERCISE  XVII 

1.  Find  sin  |  a   and  cos-|«,   if   sin  a  =  .6;    find   tan|«,  if 
tan  a  =  4. 

2.  Determine  the  functions   of  22°  30'   from  the  functions 
of  45°. 

3.  Determine  the  functions  of  15°  from   the  functions  of 
30° ;  compare  the  results  with  those  obtained  in  Exercise  XV, 
problem  4. 

4.  Determine    the    functions  of  165°   from   the   functions 
of  330°. 

5.  Obtain  the  formula  for  tan  la  by  solving  formula  [22]. 

Transform  the  first  member  into  the  second. 

6.  sin  J  a -}- cos  i  a  =  ±  Vl  +  sin  a. 
•y  7.    sin  i  a  —  cos  ^  a  =  ±  Vl  —  sin  a. 

8.    tan(7  +  £)=§eQ«  +  tana. 


U 


CH.  IV,  §33]     RELATIONS   BETWEEN  FUNCTIONS  69 


V 


•• 


2  tan-  1-tan2" 

10.    -        —  =sin«.  11.    -       — 

1  +  tan2"  l  +  tana" 

12.  tan  -  +  cot  tj  =  2  esc  «. 

13.  1  +  cot  a  cot  "  =  esc  a  cot  -. 


y   14.    tan  i  x  +  2  sin2  1  a;  cot  x  =  sin  ce. 
\/  15.    tan3  1  a?  (1  +  cot2  1  x)3  =  8  esc3  x. 

33.   Sum  and  difference  of  the  sines  and  of  the  cosines  of 
two  angles.  —  In  Art.  28,  it  has  been  showji  that 

sin  («  +  /3)  =  sin  a  cos  /3  +  cos  a  sin  /3, 
sin  («  —  £)=  sin  ct  cos  /3  —  cos  «  sin  /?, 
cos  («  +  /3)  =  cos  a  cos  /3  —  sin  a  sin  /3, 
cos  Q«  —  /3)  =  cos  a  cos  /3  +  sin  a  sin  /3. 

From  these,  by  addition  and  subtraction,  we  have 
sin  (a  +  /3)  -f-  sin  («  —  /3)  =  2  sin  «  cos  /3, 
sin  (a  +  ff)  —  sin  (a  —  /9)  =  2  cos  a  sin  yS, 
cos  («  +  /3)  +  cos  (a  —  /3)  =  2  cos  a  cos  /?, 
cos  (a  -*~  /?)  —  cos  («  —  yS)  =  —  2  sin  a  sin  /3. 

If  we  let         «  +  £  =  .<!,  and  a  —  /3  =  B, 
then  a  =    (A  +  B),  and  /3  = 


70  -  PLANE  TRIGONOMETRY  [Cn.  IV,  §  33 

Substituting  these  values  in  the  equations  above,  they 
become 

sin  A  +  sinB  =  2sin^  +  J5)cos^(^l  -  J3),      [29,  a] 

sin  A  -  sin  B  =  2  cos  \  (A  +  J5)  sin  \(A-B),  [29,  5] 
cos^l  +  cosJ5  =  2  cos  |  (^  +  J5)cos|(^  -  .B),  [29,  c] 
cos  ^  -  cos  B  =  -  2  sin  *  (^.  +  JB)  sin  \(A-B).  [29,  d] 

EXERCISE  XVIII 
Prove  the  following  identities  : 


cos  3  ct  —  cos  5  a 
sin  7  a  —  sin  a       cos  4  a 

1     ^,      -  -   -  -  • 

sin  8  a  —  sin  2  «      cos  5  a 

\          sin  a  —  sin  /?  _  tan  -%-  (ce  —  ff) 
sin  «  +  sin  ^8      tan  ^  (a  +  ^8) 


4. 


cos  5°  —  cos  47 
5.   sin  2  a  +  sin  4  «  -f  sin  6  «  =  4  cos  a  cos  2  a  sin  3  a 


6. 


cos  a  +  cos  2  a  +  cos  3  « 


sin  2  «  +  sin  2  ft  _  tan  Q  +  /?) 
sin  2  «  —  sin  2  /?  ~~  tan  («  —  @) 


8. 


cos  a  -+-  cos  (3 
9.  4^ — a      Smr  a  =  2  cos  a  —  sec  a. 

Qir>  A  /v  _L  Qin    s.  rt 


10. 


sin  4  a.  + 

sin  75°  -  sin  15°  _  V3 
cos  75°  +  cos  15°       3  ' 


CH.  IV,  §33]     RELATIONS  BETWEEN   FUNCTIONS  71 

EXERCISE   XIX 

Prove  the  following  identities  : 


1  -  tan  a  tan  ft 

2.  (sin  «  +  cos  a)2  =  1  +  sin  2  a. 

3.  cos4  «  —  sin4  a  =  cos  2  a. 

4.  tan  3  -a  —  tan  «  =  2  sin  a  sec  3  a. 

5.  cot  a  —  2  cot  2  «  —  tan  #. 

ar/"-  4-  P\  -  2  csc  2/3  +  sec/?. 
- 


7.      sin  -—  cos  -     =  1  —  sin  a. 


8.    sin  [  -  4-  a  }  sin  [  ^  —  a  )  =  \  cos  2  a. 

V4        J         V4        J 
9.14-  tan  a  tan  -  —  sec  a. 

.    tan  ( —  4-  a  ]  —  tan  (  -  —  a  }  =  2  tan  2  a. 

V4     /        V4     / 

f   11.    sin  3  a  4-  sin  2  a  —  sin  a  =  4  sin  a  cos  "  cos  — -« 

22 
IT  12.    sec2  a(l  4-  sec  2  a)  =  2  sec  2  a. 

».    esc  a  —  2  cot  2  a  cos  «  =  2  sin  a. 
14.    cos  6  a  =  32  cos6  a  —  48  cos4  a  +  18  cos2  a  —  1. 


sin  a 
esc  2  a  14-  tan2  a 


1  4-  esc  2  a      (1  4-  tan  «)2 
.    (cos  2  «  4-  cos  2  ft)2  +  (sin  2  a  4-  sin  2  /3)2  =  4  cos2  (a 


cos  «  —  cos 


.   sin  |  a  =  ±  (1  -f-  2  cos  a)-yi 


—  cos  a 


2 
20.    3  sin  2  a  —  sin  6  a  =  32  sin3  a  cos3 


CHAPTER   V 


INVERSE  FUNCTIONS  AND   TRIGONOMETRIC  EQUATIONS 


(2n+l)7T 


T 

(2  n  +  '/n)  TT 


,     34.    General  values.  —  If  sin  x  =  -,  x  =  ^,  or  — ^,  or  any 

of  the  other  angles  which  have  the  same  terminal  lines. 
There  are,  then,  an  indefinite  number  of  angles  which 
satisfy  this  equation,  or  any  similar  one.  It  is  convenient 
to  have  general  formulas  to  express  all  angles  which  have 
the  same  sine,  cosine,  or  tangent,  and  we  shall  now  pro- 
ceed to  find  such  formulas.  We  shall  first  obtain  general 
expressions  for  all  angles  which  have  their  terminal  lines 

along  one  of  the  axes.  Let  n  be 
any  integer,  positive,  negative,  or 
zero.  Then  all  angles  which 
C  have  their  terminal  lines  in  coin- 
cidence with  the  initial  line  OX 
are  represented  by  2  nir ;  for  this 
expression  represents  the  series 
of  angles  0,  2  TT,  4?r,  etc.,  and 
—  2?r,  —  4?r,  etc.,  which  evidently  contains  all  the  angles 
which  have  their  terminal  lines  in  coincidence  with  OX. 

In  like  manner  (2n  +  l)7r  represents  all  the  angles 
which  have  their  terminal  lines  in  coincidence  with  OX' ; 
for  TT  is  the  smallest  of  these  angles  and  the  addition  or 
subtraction  of  any  multiple  of  2  TT  will  evidently  give  the 

same  terminal  line. 

72 


Y' 

FIG.  47. 


CH.  V,  §  34] 


INVERSE  FUNCTIONS 


73 


Y 
FIG.  48. 


Let  the  student  show  that  the  angles  which  have  their 
terminal  lines  along  OY  are  represented  by  (2n  +  J)TT; 
and  along  OY'  by  (2w-£>7r. 

If  OP  is  the  terminal  line  of  any 
angle  «,  all  the  angles  (positive  and 

negative)  which   have   OP   as   their   x'_; i^  «**  y      ^ 

terminal  line  are  represented  by  2  WTT 

+  «.     All  angles  which   differ  from 

each  other   by   any  multiple   of   2  TT 

have    the    same   terminal    line,    and 

hence   the   same   functions.      This  fact  is  expressed   by 

saying  that  they  are  periodic  functions,  having  a  period 

Of     27T. 

Let  the  student  show  that  the  period  of  the  tangent 
is  TT. 

All  angles  ivMcTi  have  the  same  sine  or  cosecant  as  a  may 
be  expressed  by  2  mr  +  «  and  (2  n  +  1)  TT  —  «. 

For  it  was  shown  in  Art.  22  that  sin  a  =  sin  (TT  —  a) . 
Then  all  angles  which  have  the  same  terminal  lines  as 

a   and    ?r  —  a   are   expressed   by 
2  nir  +  «,  and  2  n-rr  +  (TT  —  a)  or 
(2  n  +  1)  TT  —  a.       Hence    the 
p  theorem. 

This  result  may  also  be  shown 
easily  from  Fig.  49,  where  a  is 
taken   as   an   angle   in   the    first 
quadrant.      Similar  figures  may 
be  drawn  for  any  value  of  a. 

Since  esc  a  = ,  these  are  also  the  formulas  for  all 

sin  a 

angles  which  have  the  same  cosecant. 


o 


Y 
FIG.  49. 


74  PLANE  TRIGONOMETRY  [Cn.  V,  §  34 

EXAMPLE  1.     Give  general  expressions  for  «,  if  sin  a 


The  smallest  positive  value  of  a  is  evidently  —  •     Then 
the  general  expressions  for  a  are 


2  nir  +  \  TT  and  (2  n  -f  1)  TT  -  \  TT, 
(2  n  +  1)  TT  and  (2  ^  +  f  )  TT. 

These  are  seen  to  represent  the  series  of  angles 


or 


-TT,       TT, 
—  |TT,    —  J  TT, 


TT, 


etc., 


and  —  |TT,    —  J  TT,    ---^-Tr,    ---\5-7r,  etc. 

EXAMPLE  2.     Give  general  expressions  for  a,  if  sin  a 

=-f 

The  smallest  positive  value  of  a  is  ^  ?r.     Then  the  gen- 
eral expressions  for  a  are 

2  mr  +  I  TT  and  (2  n  +  1)  TT  -  -J  TT, 
or  (2/i  +  I)  TT  and  (2  w  -  -J)  TT. 


angles  which  have  the  same  cosine  or  secant  as  ct  may 
be  expressed  by  2  mr  ±  a. 

For  it  was  shown  in  Art.    22  that  cos  a  =  cos  (—  a). 
Then   all    angles   which    have    the    same    terminal   lines 

as  a  and   —  a  are  expressed   by 
2  mr  +  «    and    2  nir  —  a  ;     or    in 
one   formula,    2  mr  ±  a.      Hence 
~      the  theorem. 

This  result  may  also  be  shown 


X- 


FIG.  50. 


easily  by  the  aid  of  Fig.  50. 


CH.  V,  §  34] 


INVERSE  FUNCTIONS 


75 


EXAMPLE.  Give  the  general  expression  for  «,  if 
cos  a  =  —  J.  The  smallest  value  of  a  is  evidently  |  TT. 
Then  the  general  expression  for  a  is  2  UTT  ±  f  TT,  or 
(2*±f>r. 

J.ZZ  angles  which  have  the  same  tangent  or  cotangent  as  a 
may  be  expressed  by  mr  -f  #• 

For  it  was  shown  in  Art.  22  that  tana  =  tan(?r+  a). 
All  angles  which  have  the  same  terminal  lines  as  a  and 
7r+«  are  expressed  by  2n7r  +  a  and  2njr  +  (TT  +  a),  or 
(2  M  -f-  1)  TT  +  a.  But  since  2  w  and  2^  +  1  include  all 
integers,  these  two  formulas  may  be  written  as  the  one, 
TITT  +  a.  Hence  the  theorem. 

This  result  may  also  be  shown  by  the  aid  of  Fig.  51, 
since  it  is  evident  that  all 
angles  which  have  the  same 
tangent  have  their  terminal 
lines  in  the  same  line  through 
the  origin.  We  shall  evidently 
reach  one  or  the  other  of  these  FlG-  51- 

terminal  lines  when  we  add  any  multiple  of  TT  to  a. 


EXERCISE  XX 

Give  general  expressions  for  x,  if 

1.  sinx  — 0.  8.  sec#  =  l. 

2.  cos  x  =  0.  9.  sec  x  =  — 1. 

3.  tan  x  =  0.  10.  tan  x  =  1. 

4.  sina?  =  l.  11.  tan#  =  —  1. 

5.  sin  05  =  —  1.  12.  sin  x  =  - 

6.  cos  05  =  1.  13.  sin#  = 

7.  cos  x  =  — 1.  14.  cos  x  =  • 


7G  PLANE  TRIGONOMETRY  [Cn.  V,  §  35 

15.  cos#  =  jV2.  ^X~23.  cos-aj=l. 

16.  cos  x  —  —  \  V2.  24.  sin  x  =  1. 

17.  sin  x  =  ^  V3.  25.  tan  x  =  2. 
4^18.  eosa?  =  — JV3.  26.  cosx  =  —  -J. 
^19.  tan2  a?  =  i.  27.  sec  a;  =  10. 
^20.  cos-a  =  -J.  28.  esc2  a;  =  2. 

21.  tan  a  =  2 -VI  29.    tan2  a?  =  3. 

22.  secx  =  —  2.  30.    tan2  a;  =  7  —  4  V3. 

35.  Inverse  trigonometric  functions.  —  Throughout  all 
the  previous  work  the  trigonometric  ratios  have  been  con- 
sidered as  functions  of  the  angle ;  but  it  is  also  possible 
to  think  of  the  angle  as  a  function  of  the  trigonometric 
ratios.  For  this  purpose,  if  y  —  sin  #,  it  is  convenient  to 
have  a  short  method  of  writing  the  fact  that  "  x  is  an 
angle  whose  sine  is  y"  The  usual  method  employed  in 
this  country  is  x  =  s'm~ly,  which  may  be  read  "a;  equals 
anti-sine?/,"  or  "inverse-sine  y";  but  the  student  must 
remember  that  this  is  only  an  abbreviation  for  the  longer 
statement  "a;, is  an  angle  whose  sine  is  ?/."  With  the 
same  meaning,  we  use  x  =  cos"1  y,  x  =  tan"1  y,  x  —  sec"1  y, 
etc.  These  are  called  inverse  trigonometric  or  inverse 
circular  functions. 

We  have  seen  that  when  an  angle  is  given,  its  trigo- 
nometric functions  are  determined.  For  example,  if 
y  =  cos  a;,  y  has  a  single  determinate  value  for  every 
given  value  of  x.  But  if  a  value  is  given  to  y,  it  has 
been  shown  in  the  previous  article  that  x  has  an  indefinite 
number  of  values.  Thus,  if  y  =  -J-,  x  =  cos"1  \ ?  =  2  nir  ±  ^« 

o 

We  then  may  use  x  =  cos"1 1  and  x  =  2  mr  ±      as  different 


CH.  V,  §  35]  INVERSE   FUNCTIONS  77 

methods  of  expressing  the  same  idea.  But  in  working 
with  inverse  trigonometric  functions  we  shall  understand 
that  the  least  positive  angle  of  the  set  is  meant,  unless  the 
contrary  is  stated.  This  least  positive  value  is  called  the 
principal  value  of  the  function. 

With  this  meaning  we  may  prove  the  equality  of  such 
expressions  as  sin"1  x  and  cos"1  Vl  —  x2.  But  these  two 
expressions  do  not  represent  the  same  series  of  angles,  as 
may  readily  be  seen  by  giving  x  a  numerical  value.  If 
x  =  £,  Vl  -  x2  =  I V3,  and 

sin-1  a;  =30°,  150°,  390°,  510°,  etc., 
while     cos-1  Vr^?  =  30°,  330°,  390°,  690°,  etc. 

EXAMPLE  1.     Prove  that  sin"1  x  =  cos"1  Vl  —  x2. 

Let  y=*sin  OJ.  Then  x  =  s\ny.  From  this  we  see 
that  cos  y  =  Vl  —  x2.  Hence  y  =  cos"1  Vl  —  x2,  and 
sin"1  x  =  cos"1  Vl  —  x2. 

EXAMPLE  2.     Prove  that  tan"1  x  +  tan-1  y= tan-1  ^ 

1-xy 

Let  tan"1  x  =  z  and  tan ~l  y  =  w. 
Then  x  =  tan  z  and  y  =  tan  w. 

T)   ,  N        tan  z  +  tan  w        x  +  y 

But        tan  (z  +  w)  =  —  —  =         y  . 

1  —  tan  z  tan  w      \  —  xy 

Hence  z  +  w  —  tan"1  ?     ^ , 

\—xy 

and  tan"1  x  +  tan"1  y  =  tan"1  ^     ^  . 


78  PLANE  TRIGONOMETRY  [Cn.  V,  §  35 

1  —  r2 

EXAMPLE  3.     Prove  that  2  tan-1  x  =  cos"1  —  — '-• 
Let  tan"1  x  =  y.     Then  x  =  tan  ?/. 

"1  9 

We  wish  to  show  that  2  y  =  cos"1     ~  x  - 

But  cos  2  y  =  cos2'?/  —  sin2  ?/, 

_         1  tan2?/ 

~  1  +  tan2  y      1  - 


Hence  2 «/  =  cos"1  —  — ^ 

"I  9 

and  2  tan"1  x  =  cos"1  — -  — - 

EXERCISE  XXI 

Prove  the  following  identities  for  the  principal  values  of  the 
inverse  functions. 

1.  sin  (cos-1 1)  =  1 V3.  3.    tan  (2  tan"1  £)  =  f . 

2 .  tan  (sin-1  if)  =  -L2 .  4.    tan  (£  ta.n-1  if)  =  |. 

5 .  tan  (tan-1  J  —  tan-1 1)  =  i. 

6.  cos"1  x  =  sin"1  Vl  —  x2. 

8.  2  cos-1  x  =  sin-1  (2  a;  Vl  -  x2). 

9.  tan^i-f  tan-1i  =  -. 

10.   3  tan-1  x  =  tan-1  — 


CH.  V,  §  30]  TRIGONOMETRIC  EQUATIONS  79 

11.   tan(2tan-1a)=-^-0. 
I  -a2 


2i  —  a 

14.  sin  (sin-1  a  +  sin-1  6)  =  a  Vl  -  6*  +  6  Vl  -  a2. 

15.  tan-1  1  +  tan-1  1  +  tan-1  1  +  tan-1  f  =  -• 

36.  Solution  of  trigonometric  equations.  —  In  all  the 
previous  work,  except  Art.  34,  the  equations  with  which 
we  have  dealt  have  been  true  for  all  values  of  the  angles, 
and  the  student  has  been  asked  to  prove  this  fact.  There 
is  another  important  class  of  problems  in  which  the 
student  is  given  an  equation  which  is  true  only  for  cer- 
tain values  of  the  angle,  and  he  is  asked  to  determine  the 
values  for  which  it  is  true.  In  other  words,  he  is  asked 
to  "  solve  the  equation."  The  first  step  toward  this  end 
is  usually  to  transform  the  given  equation  into  one  which 
contains  a  single  trigonometric  function  of  a  single  angle. 
This  function  may  then  be  looked  upon  as  the  unknown 
quantity  and  its  value  may  be  obtained  by  the  algebraic 
solution  of  the  equation  for  this  unknown.  If  the  equa- 
tion can  be  reduced  to  either  a  simple  or  a  quadratic 
equation,  the  solution  may  be  obtained  by  elementary 
methods,  and  only  such  equations  will  be  considered  here. 
The  following  problems  illustrate  the  method  of  pro- 
cedure in  the  simpler  cases,  where  the  equation  contains 
functions  of  a  single  angle  only. 


80  PLANE  TRIGONOMETRY  [Cn.  V,  §  36 

EXAMPLE  1.     Solve  the  equation  sin  x  +  esc  x=2. 

Since  esc  x  =  -  ,  the  equation  becomes 

sinx 

sin  x  -\  --  =  2. 


Clearing,  sin2  x  —  2  sin  x  -f-  1  =  0. 
Solving,  sin  x  =  1, 

x  =  sin"1  1, 


*=!• 


All  values  of  #  are  then  represented  by 


EXAMPLE  2.     Solve  the  equation  sin  #  =  tan2  a?. 

Since  tana;  =  5HLE,  the  equation  becomes 

COS  X 


sin  x  = 

cosa; 

Clearing,  sin  #  cos2  a;  =  sin2  x, 

or  sin  x  (cos2  #  —  sin  x)  =  0. 

Hence  sin  x  =  0, 

and  cos2  #  —  sin  #  =  0. 

The  first  of  these  equations  gives  at  once 

x  =  0,  or  TT, 

=  717T. 


CH.  V,  §  37]  TRIGONOMETRIC   EQUATIONS  81 

The  second  equation,  cos2  x—  sin  x=  0,  may  be  written 
1  —  sin2  x  —  sin  x  —  0. 

Solving  as  an  affected  quadratic  in  sin  a?,  we  have 
sin  x  =  ~  l  *  V^=  0.61803,  or  -1.61803. 

The  second  of  these  values  is  evidently  impossible, 
since  sin  a;  cannot  be  numerically  greater  than  1.  Im- 
possible solutions  of  this  nature  are  of  frequent  occurrence 
in  solving  trigonometric  equations.  This  value  of  sin  a; 
satisfies  the  original  equation,  but  since  there  is  no  angle 
whose  sine  is  —  1.61805,  it  does  not  give  a  possible  solu- 
tion of  that  equation. 

From  the  other  value  of  sin  a;,  we  have  x  =  sin"1  .61803 
=  38°  10'  21".  There  will  also  be  an  angle  in  the  second 
quadrant,  180°  -  38°  10'  21".  The  general  answers  are 
then  nir,  2nir  +  38°  10'  21",  and  (2  n  +  I)TT  -  38°  10'  21". 

EXERCISE  XXII 

1  .    cos2  x  =  sin2  x.  6.    sec2  a?  =  4  tan  x. 

2.  2  cos  x  =  V3  cot  x.  7.    tan2  x  —  sec  x  =  1. 

3.  tan  x  -f  cot  x  =  2.  8.    tan2  x  +  esc2  x  =  3. 

4.  sec  x  +  2  cos  x  =  3.  9.    4  tan  x  —  cot  x  =  3. 

5.  sinic  =  tan#.  10.    tan2  x  +  cot2  x  =  *-. 


37.  There  are  many  equations  in  which  it  is  convenient 
to  introduce  expressions  containing  radicals  when  we 
attempt  to  write  the  equation  in  terms  of  a  single  func- 
tion. It  will  then  be  necessary  to  square  the  equation 
in  the  solution;  and  in  doing  this  an  ambiguity  will  be 


82  PLANE  TRIGONOMETRY  [Cn.  V,  §  37 

introduced,  and  the  solutions  of  the  resulting  equation 
will  contain  not  only  the  solutions  of  the  original  equa- 
tion, but  also  the  solutions  of  the  equation  obtained  by 
changing  the  sign  before  the  radical.  It  then  becomes 
necessary,  as  in  any  radical  equation,  to  determine,  by 
actual  substitution,  which  of  the  results  satisfy  the  given 
equation.  This  difficulty  may  often  be  avoided  by  using 
formulas  which  do  not  contain  radicals,  but  it  is  not 
always  convenient  to  do  this. 

EXAMPLE.     Solve  the  equation  V3  sin  x  —  cos  x  =  1. 


Replacing  smx  by   Vl  —  cos2  2-, 


V3  Vl  —  cos?x  =  1  4-  cos  a;. 

Squaring,         3  (1  —  cos2#)  =1  +  2  cosx  +  cos2#. 
Uniting,  4  cos2 x  +  2  cos  x  —  2  =  0. 

Solving,  cos#=|,  or  —1. 

,    7T 

x  =  ±  — ,  or  TT. 
o 

If   cos  x  —  J,   sin  x  —  ±  |  V3,    and  the  equation  is  evi- 
dently satisfied  if  sin  x  =  -f-  \  A/3,  but  not  if  sin  x  =  —  |  A/3. 

The  solution   —  ~  must,   therefore,  be   discarded.     It  is 
o 

easily  seen  that  the  equation  is  satisfied  when  x  =  IT. 
The  general  solutions  are,  then,  (2  n  +  J)TT,  and  (2  n-\-V)ir. 
SECOND  SOLUTION.  —  This  difficulty  may  be  avoided  in 
any  problem  which  is  in  the  form  a  sin  x  +  b  cos  x  =  c  by 
the  following  device :  Divide  the  equation  through  by 

Va2  +  b2.     Then  —   a         may  always  be   expressed   as 

Va2  +  V  b 

the  sine  of  some  angle,  and  ~   as  the  cosine  of 


CH.  V,  §  38]  TRIGONOMETRIC   EQUATIONS  83 

the  same  angle.     In  this  problem  Va2  +  62  =  2.     Dividing 

V3   .  1  1 

by  2, 


Substituting  —  —  =  sin^,  and  -=  cos^,  we  have 

L  O  2  O 

sin—  sinx  —  cos—  cosx  =  -,  or  cos(^-f#)=  --  . 

2  \3        /         2 

Hence 

±7r,  and  z=(2ra-l)7r,  or 


EXERCISE  XXIII 

1.  sin  aj  —  cos  a?  =  0.  **B.    cot  x—  tan  x=  since  +  cos  x. 

2.  sinic+  cosic  =  1.  If  fl.    cos  cc  +  tan  x  =  sec  a?. 

3.  sin  a?  —  cos  x  =  —  V|.  ^"8.    esc  x  =  1  +  cot  x. 

4.  esc  a/*  —  cot  x  =  V3.  9.    tan  a?  +  sec  x  =  VS. 
^5.    ^  cos  x  —  J  sin  x  =  ^.              10.    5  sin  a;  -f-  2  cos  x  =  5. 

38.  In  the  previous  problems  only  functions  of  x  have 
occurred.  If  the  equation  contains  functions  of  multiples 
of  #,  as  2#,  3#,  J#,  etc.,  these  may  all  be  replaced  by 
their  values  in  terms  of  functions  of  x,  and  the  equation 
solved  as  in  the  previous  problems.  But  many  equations 
may  be  solved  more  readily  by  various  devices,  and  some 
may  be  solved  by  these  devices  which,  if  expressed  in 
functions  of  #,  would  give  equations  of  the  third  or 
higher  degree,  which  the  student  could  not  solve.  There 
is  an  excellent  chance  for  the  display  of  the  ingenuity 
of  the  student  in  discovering  methods  for  shortening  the 
work  in  many  of  these  problems.  A  few  of  these  are 
illustrated  in  the  following  examples. 


84  PLANE  TRIGONOMETRY  [Cn.  V,  §  38 

EXAMPLE  1.     Solve  the  equation 

sin  3  x  +  sin  2  x  +  sin  x  =  0. 

By  formula  [29,  a], 

sin  3  x  +  sin  a;  =  2  sin  2  a;  cos  x. 

The  given  equation  may  then  be  written 

sin  2  x  =  —  2  sin  2  a;  cos  #. 
From  which      sin  2  a;  =  0,    and    cos  x  =  —  J. 
Hence  2  z  =  WTT,  #  =  (2  n  ±  f )  TT. 

*  =  T* 

The  values  of  x  less  than  360°  are  seen  to  be  90°,  120°, 
180°,  240°,  270°. 

EXAMPLE  2.     Solve  the  equation 

esc  x  —  cot  x  =  V3. 
By  substitution  this  becomes 


or 


COS  X 

sin  x      sin  x 

1  —  cos  x 
sin  a; 


=  V3, 


VS. 


This  reduces,  by  formula  [25,  5],  to 
tan  |-  #  =  VS. 

Hence  J  #  =  nir  4-  ?» 

o 

and  a;=(2™  +  f)7r. 

The  only  value  of  x  less  than  360°  is  seen  to  be  120C 


OIL  V,  §  38]  TRIGONOMETRIC   EQUATIONS  85 

EXAMPLE  3.     Solve  the  equation  cos  3  x  =  cos  2x. 

From  Art.  34,  cos  2  x  =  cos  (2  mr  ±  2  x). 
Hence  3  x  =  2  nir  ±  2  #, 

and  #  =  2  WTT,    and    5  x  =  2  WTT,    or   a?  =  |  HTT. 

The  values  of  ^  less  than  360°  are  seen  to  be  0,  72°, 
144°,  216°,  288°. 

EXAMPLE  4.     Solve  the  equation 

2  sin  x  sin  3  x  =  1. 

By  formula  [29,  d~]  this  becomes 

cos  2  x  —  cos  4  a;  =  1. 
By  formula  [18,  <?], 

cos2#-  (2cos22#—  1)  =  1, 

or  2  cos2  2  #  =  cos  2  x. 

Solving, 

cos  2  a;  =  0,  and    cos  2  x  =    . 


Hence    2z=,  , 

2  o 


and  a;=     WTT±J,        or  x—     n-rr  ±^. 

The  values  of  x  less  than  360°  are  seen  to  be  30°,  45°, 
135°,  150°,  210°,  225°,  315°,  330°. 


86  PLANE  TRIGONOMETRY  [Cn.  V,  §  38 

EXAMPLE  5.     Solve  the  equation 

sec  x  =  2  (sin  x  +  cos  x). 

Replacing  sec  a;  by ,  and  reducing,  we  have 

cos  a; 

1  =  2  sin  x  cos  x  +  2  cos2 re. 


This  might  be  solved  by  replacing  sin  x  by  VI  —  cos2 re, 
but  we  have  seen  that  it  is  best  to  avoid  the  introduction 
of  radicals.  But  if  we  notice  that  2  sin  x  cos  x  —  sin  2  re, 
and  2  cos2  re  —  1  =  cos  2  re,  the  equation  becomes 

sin  2  x  +  cos  2  x  =  0. 
Dividing  by  cos  2  x,      tan  2  x  =  —  1. 
Hence  2  re  =  WTT  +  f  TT, 

-,  WTT  .    o 

and  re  =  — -  +  |  TT. 

EXERCISE  XXIV 

1.  sin  2 re  =  2 cos  re.  9.    cos 2 re—  sinre  =  £. 

2.  4  cos  2  re +  3  cos  re  =  1.  10.    tan  2  re  tan  3  re  =  1. 
Vs.    tan  2  re  =  tan  re.                        11.    2  sin2  re  +  sin2  2  re  =  2. 

4.  2  sin2  2  re  =  1  —  cos  2  re.  12.  sin  re  +  cos  re  +  sin  2  re  =  2. 

5.  cos  3  re  —  cos  5  re  =  sin  re.  13.  tan  re  =  4  sin  1  re. 

6.  cos  5  x + cos  3  re + cos  re =0.  14.  tan  re  —  tan  3  re  =  2. 

7.  2  cos  re  cos  3  re +  1  =  0.  15.  cot  re  —  esc  2  re  ==  1. 

8.  tan  3  x  =  3  tan  re.  16.  cos  3  re  +  2  cos  re  =  0. 


CH.  V,  §  38]  TRIGONOMETRIC   EQUATIONS  87 

EXERCISE  XXV.    REVIEW 

£   1.    If  sec  (9  =»  3f ,  find  the  value  of  tan  -• 

2 

K   2.    If  cot  a  =  2  —  V3,  prove  that  sec  a  =  V6  -f-  V2,  and  that 
esc  a  =  V6  —  V2,  numerically. 

I/    3.  Prove  that  cos  (135°  +  A)  +  cos  (135°  -  A)  =  -  V2  cos  -4. 

4.  Prove  that  (sec  15°  +  esc  15°)2  =  24. 

5.  Find  the  value  of  sin  15°  +  cos  15°. 
Vs.  Prove  that  tan  75°  -f  cot  75°  =  4. 

7.  Find  the  value  of  tan  105°. 

8.  Prove  that  tan  60°-  tan  165°=  2. 

9.  If  vers  A  =  -,  find  tan  —  • 

10.  Prove  that  sin  [  -  +  6  }  =  cosf  -  _  0\ 

V4        )  \±       ) 

11.  If  tan ^4=        3  _,  and  tan  B  =  _,  find  tan  (A-B\ 

4-V3  4  +  V3 

12.  Prove  that  cos  20°  +  cos  100°  +  cos  140°  =  0. 

13.  The  cosines  of  two  of  the  angles  of  a  triangle  are  f  and 
-f%  respectively ;  find  the  tangent  of  the  third  angle. 

14.  Solve  the  equation  sin  x  -\-  cos  x  =  sec  x. 

15.  Construct  geometrically  an  angle  whose  secant  is  —  3. 

16.  Prove  that  cos2  A  +  cos2  (A  + 120°)  +  cos2  (A  - 120°)  =  f . 

17.  Prove  that  vers  (270°  +  A)  vers  (270°-  A)=  cos2  A. 

18.  Solve  the  equation  cot2 0  —  tan2  0  =  2  esc  9  sec  9. 

19.  Prove  that  2  vers(|  +  ^  versf|~l)=  yers(7r-  e)' 

20.  Prove  that  4  sin  20°  sin  40°  sin  80°  =  sin  60°. 


88  PLANE  TRIGONOMETRY  [Cn.  V,  §  38 

21.  Prove  that  sin"1-  +  sin-1—  +  sin"1—  =  -. 

5  13  65      2 

22.  If  sin  A  =    0        ,  and  sin  B  =    Jp  ,  find  tan  (A  +  E). 

n2  +  1  2  +  1 


23.  Express  in  terms  respectively  of  the  secant  and  cosecant 
of  A  and  B  :  (1)  sec  (A  +  5),  (2)  esc  (A-'B). 

24.  Prove  that  sec  105°  =  -  V2  (V3  +.1). 

25.  Prove  that  tan^"2^1  +  tan^±_^d  =  2  sec  A 

4  4 

26.  Prove  that  esc  2  a  +  cot  2  «  =  cot  «. 

27.  Solve  the  equation  sin2  2x  —  sin2  x  =  \. 

28.  Prove  that  sec-1  3  =  2  cot'1  V2. 

29.  Prove  that  tan"1^  +  tan^l  +  tan'1^  +  tan-1^  =  £. 

o  o  7  o       4 

30.  If  tan  A  +  tan  2  ^4  =  tan  3  A,  prove  that  yl  must  be  a 
multiple  of  60°  or  90°. 


CHAPTER    VI 


OBLIQUE  TRIANGLES 

39.  We  shall  now  proceed  to  prove  three   theorems, 
connecting  the  sides  and  angles  of  a  triangle,  which  will 
enable  us  to  solve  any  triangle.     In  every  case  let  the 
triangle  be  lettered  ABC,  and  let  a,  b,  c  represent  the 
lengths  of  the  sides  opposite  the  corresponding  angles. 
In  these  proofs  no  account  is  taken  of  the  positive  or 
negative  direction   of  the   sides  or  angles.     The  letters 
a,  5,  and  c  simply  represent  the  positive  magnitudes  of 
the  sides,   and  A.,   B,  and    C  the  interior  angles  of  the 
triangle.      These  forms  of   the  theorems  are,  therefore, 
only  suited  to  the  solution  of  triangles ;  they  cannot  be 
used  when  the  directions  as  well  as  the  magnitudes  of  the 
sides  and  angles  need  to  be  considered.     For  the  general 
form  of  these  theorems  and  their  proof,  see  Art.  43. 

40.  Law  of  the  sines. — In  any  triangle,  the  sides  are 
proportional  to  the  sines  of  the  opposite  angles. 

G 


A 


D 
FIG.  52  (a). 


B 

FIG.  52  (6). 


89 


90  PLANE  TRIGONOMETRY  [Cn.  VI,  \  41 

In  either  Fig.  52  (a)  or  52  (5),  let  the  length  of  the 
perpendicular  DO  be  represented  by  h.  Then  in  trian- 
gle A  CD, 


In  triangle  D  CB,     sin  B  =  -• 

a 

(In  Fig.  52  (6),       -  =  sin  (TT  -  B)  =  sin  £. 
a 

Hence  by  division, 


By  drawing  perpendiculars  from  the  other  vertices,  the 
same  relations  may  be  shown  to  hold  between  any  pair 
of  angles  and  sides. 

b      sin  B     a      sin  A 


c      sin  O     c      sin  (7 
These  three  formulas  may  be  written  in  the  form 


, 

sin  A     sin  B     sin  O 

where  r  may  be  shown  to  be  the  radius  of  the  circum- 
scribed circle. 

41.  Law  of  the  cosines.  —  The  square  of  any  side  of  a 
triangle  is  equal  to  the  sum  of  the  squares  of  the  other  two 
sides  diminished  by  twice  the  product  of  these  sides  and  the 
cosine  of  the  included  angle. 

Since  we  are  making  no  use  of  the  directions  of  the 
sides,  the  proof,  when  the  included  angle  is  obtuse,  will 


CH.  VI,  §  41] 


OBLIQUE   TRIANGLES 


91 


differ  slightly  from  the  proof  when  the  angle  is  acute. 
It  seems  best,  therefore,  to  give  separate  proofs  of  the 

two  cases. 

c 


A  D 

FIG.  53  (a). 


U 


FIG.  53  (6). 

In  Fig.  53  (a),  BO2  =  CD2  +  (AB  -  AD)*, 

=  CD2  +  AS*  +  AI?  -  2  AB-AD. 
But          CD2  +  AD2  =  AC2,  and  AD  =  AC  cos  A. 
Hence  a2  =  b2  +  c2  —  2  be  cos  A. 

In  Fig.  53  (5),  BC2  =  CD2  +  (DA  + 


But         Off  +  DA2  =  AC2, 

and  DA  =  AC  cos  CAD  =  -  AC  cos  A. 

Hence  a2  =  b2  +  c'*  -  2  be  cos  A.  [31] 

We  see  then  that  the  formula  holds  for  sides  opposite 
either  acute  or  obtuse  angles.  In  fact,  the  first  proof  is 
sufficient,  if  account  is  taken  of  the  directions  of  the  lines 
AB  and  AD. 

By  dropping  perpendiculars  from  the  other  vertices, 
similar  expressions  may  be  found  for  the  other  sides. 

62  =  c2  +  a2  -  2  ca  cos  J5, 
c*  =  a?  +  b*-2ab  cos  (7. 


92  PLANE  TRIGONOMETRY  [Cn.  VI,  §  42 

42.    Law  of  the  tangents.  —  The  sum  of  any  two  sides  of 
a  triangle  is  to  their  difference  as  the  tangent  of  one-half 
of  the  sum  of  the  opposite  angles  is  to  the  tangent  of  one- 
half  their  difference. 
From  formula  [30], 

a_  _  sin  .A 
b      sin.Z? 

By  composition  and  division,  this  proportion  becomes 

a  +  b  _  sin  A  +  sin  B 
a  —  b      sin  A  —  sin  B 

But,  by  formula  [29,  a~]  and  [29,  J], 

sin  A  +  sin  B  _  2  sin  1  (  A  +  B)  cos  \(A  -  B) 
sin  A  -  sin  B  ~  2  cos  1  (A  +  ^)  sin  l(A-  .#)' 


tan 

«46     tan    Ql  +  J?) 

Hence  -  —  v=-  —  .  [32] 

a~6 


In  like  manner,  it  may  be  shown  that 
b  +  c  _  tanl(^+  C) 


c  ~~  tai      (5  -  Oy 

a  _  tan|((74-  J.) 
== 


c-atani(C'-  A) 

43.*  General  form  of  the  law  of  the  sines  and  the  law  of 
the  cosines.  —  In  the  three  laws  just  obtained  only  the 
magnitude  of  the  sides  and  of  the  interior  angles  have 
been  considered.  No  attention  has  been  paid  to  the 


CH.  VI,  §  43] 


OBLIQUE   TRIANGLES 


93 


signs  of  either  the  sides  or  angles.  But  in  using 
either  the  law  of  the  sines  or  the  law  of  the  cosines  in 
other  branches  of  mathematics,  the  sides  and  angles  of 
the  triangles  are  often  directed  lines  and  angles,  and  it 
is  convenient  to  have  a  form  of  these  laws  which  will 
apply  to  such  cases. 

Law  of  the  sines. 

Let  the  sides  of  the  triangle  ABO  be  directed  lines. 
If  the  positive  direction  of  A  C  is  from  A  to  (7,  place  the 


*r-X 

IS 

FIG.  54  (a). 

triangle  with  A  at  the  origin,  as  in  Fig.  54  (a) ;  if  its 
positive  direction  is  from  C  to  A,  place  it  as  in  Fig. 
54  (t). 

In  either  figure, 

DB  =  projyAB  =  proj^  CB. 
Hence,  by  [10], 

AB  sin  (AC,  AB)  =  CB  sin  (AC, 


or 


AB 


CB      sin(AC,AB) 


or,  changing  the  sign   of   the   denominator  of   the   first 
member  and  of  the  numerator  of  the  second, 


BC     sin  (6,  c) 


[33] 


94  PLANE  TRIGONOMETRY  [On.  VI,  §  43 


In  like  manner,  =  ',  etc. 

OA      sin  (c,  a) 

Law  of  the  cosines.  —  Placing  the  triangle  as  in  the  pre- 
vious demonstration, 

in  either  figure,          DA  =  DO  +  OA. 

Squaring,  DA*  =  DO*  +  OA2+2DO.OA. 

Adding    Dl?,    and    noting    that    DA2  +  BD*  =  A&, 
=  BC\  and  that 


DO=pvojx£0=£Ocos  (OA,  BO)  =  BO  cos  (BO,  OA), 
we  have    A&  =  BC2  +  CA1  +  2  BC  •  C^  cos(BC,  C^.).    [34] 
In  like  manner 

=  CA2  +  ^S2  +  2  (M  -  ^^  cos  (OA,  ^1^),  etc. 


In  any  numerical  case  these  laws  will  be  found  to  give 
the  same  results  as  the  special  forms  given  in  the  previous 

articles.  To  illustrate,  consider 
the  triangle  in  Fig.  55,  in  which 
the  directions  of  the  sides  are 
shown  by  arrows,  and  the  lengths 
of  the  sides  are  2,  3,  and  4.  Let 
the  letters  A,  .Z?,  and  C  represent, 
-A  K  as  usual,  the  magnitudes  of  the 

FIG.  55. 

interior   angles   of   the   triangle. 

Then  AB  =  2,  BC  =  3,  OA  =  -  4,  (BC,  CA)  =  Z.MOL 
=  -  C,  (OA,  AB)=  ^LAK=  -  A,  (AB,  £0)= 


CH.  VI,  §  45]  OBLIQUE   TRIANGLES  95 

r™     ,          P  , ,  .        2      sin(—  C)      sin  (7 

The  law  of  tne  sines  gives  -  =  —  —  =  — — -A- 

6      sin  (—  A)      sin.  A 

The  law  of  the  cosines  gives 

22  =  32  +  (_  4)2  +  2(3)(-  4)  cos  (-  <7), 
=  32  +  42  _  2  .  3  .  4  cos  <7. 

44.  Solution  of  oblique  triangles.  —  The  three  sides  and 
the  three  angles  of   a  triangle  are  spoken  of  as  the  six 
parts  of  the  triangle.     It  is,  in  general,  possible  to  find 
the  remaining  parts,  when  any  three  of  these  are  known, 
if  one  of  the  known  parts  is  a  side.     The  process  of  finding 
the  three  unknown  parts  is  called  solving  the  triangle. 

Four  cases  are  distinguished.     There  may  be  given 

(1)  two  angles  and  one  side, 

(2)  two  sides  and  the  included  angle, 

(3)  two  sides  and  an  angle  opposite  one  of  them, 

(4)  three  sides. 

45.  CASE  I.     Given  two  angles  and  one  side.  —  Let  A,  B, 

and  a  be  the  known  parts ;  it  is  required  to  find  (7,  6,  and  c. 
The  third  angle  0  is  determined  at  once  from  the  equation 

(7  =180° -04  +  5). 
In  the  law  of  the  sines, 

a  _  sin  A          c  _  sin  0 
b      sin.Z?          a      sin  A 

three  of  the  four  parts  are  known  ;  the  fourth  part  (b  or  c) 
may  be  found  at  once  by  the  aid  of  the  table  of  natural 
sines. 

The  accuracy  of  the  result  may  be  tested  by  obtaining 

j*-,  i-  mi        ^ 


96  PLANE  TKIGONOMETKY  [Cn.  VI,  §  45 

one  of  the  given  parts  from  the  three  which  have  just 
been  found,  by  the  aid  of  a  formula  which  has  not  been 
previously  used. 

EXERCISE  XXVI 

Find  the  remaining  parts  of  the  triangle,  when 

1.  A  =  7°  21',  £  =  50°  30',  a  =  9. 

SOLUTION. 

C  =  180°  -  (7°  21'  +  50°  30)  =  122°  9'. 

,  _  a  sin  B  _  9  x  .7716  _  r .  9g 
=  sin^l  =       .1279 

_  a  sin  C  __  9  x  .8467  _  ^  ^ 
=  smA~      .1279 

2.  A  =  82°  22',  B  =  43°  20',  a  =  4.79. 

3.  B  =  10°  12',  C  =  46°  36',  6  =  5. 

4.  A  =  12°  49',  (7  =  141°  59',  a  =  82. 

5.  A  =  99°  55',  5  =  45°  1',  a  =  804. 

6.  £  =  4°  20',  C  -  136°  14',  b  =  51. 

SOLUTION.  —  Logarithms  may  be  used  to  advantage  in  apply- 
ing the  law  of  the  sines. 

A  =  180°  -  (4°  20'  +  136°  14')  =  39°  26'. 

a  =  _|1B — ?  or  log  a  =  log  6  +  log  sin  A  —  log  sin  B. 
sin  jB 

c  =  —   — — ,  or  log  c  =  log  b  +  log  sin  C  —  log  sin  B. 
sin  _o 

log&=  1.70757  log&=  1.70757 

log  sin  A  =  9.80290  log  sin  (7  =  9.83993 

11.51047  11.54750 

log  sin  B  =  8.87829  log  sin  J5  =_  8.87829 

loga=  2.63218  log  c  -  2.66921 

a  =  428.73.  c  -  466.88. 


CH.  VI,  §  46]  OBLIQUE   TRIANGLES  97 

7.  A  =  38°  21'  47",  B  =  54°  6'  8",  a  =  13.509. 

8.  B  =  125°  53'  52",  C  =  15°  44'  21",  c  =  5.904. 

9.  5  =  44°  22'  49",  C=  106°  11'  53",  a  =  1879.4. 

10.  A  =  41°  13'  22",  C  =  71°  19'  5",  a  =  55. 

11.  A  =  48°  24'  15",  C  =  31°  13',  c  =  926.74. 

12.  5  =  16°  21\18",  C=  24°  17',  b  =  43.24. 

46.    CASE  II.    Given  two  sides  and  the  included  angle.— 
Let  a,  b,  and  C  be  the  known  parts ;  it  is  required  to  find 
A,  B,  and  c. 

The  law  of  the  cosines  may  be  used  to  find  the  third  side, 
if  the  given  sides  are  expressed  in  numbers  which  ma}7-  be 
easily  squared.  The  angles  may  then  be  found  by  the 
law  of  the  sines,  as  in  the  previous  case. 

EXERCISE  XXVII 

Find  the  third  side  of  the  triangle,  when 
1.   A  =  31°,  b  =  6,  c  =10. 


SOLUTION.      a  =  -vW  +  c2  —  2  be  cos  A, 

=  V36  +  100  -  120  x  .8572, 
=  5.756. 

2.  (7=50°,  a  =  10,  6  =  11. 

3.  A  =  60°,  b  =  8,  c  =  15. 

4.  C=135°,  a=V3-l,  6=V2. 

5.  £  =  30°,  a  =  3,c  = 


98  PLANE  TRIGONOMETRY  [Cn.  VI,  §  47 

47.  Solution  by  logarithms.  —  When  the  numbers  which 
express  the  length  of  the  sides  of  the  triangle  contain  a 
number  of  figures,  and  it  is  necessary  to  find  the  angles, 
the  method  given  above  is  long,  since  the  formula  is  not 
adapted  to  the  use  of  logarithms. 

For  the  general  treatment  of  this  case  by  the  aid  of 
logarithms,  the  law  of  the  tangents  is  best  suited.  It 
may  be  written 

tan  I  (A  -  E)  =  —^  tan  J  (A  +  B). 

The  second  member  of  this  equation  is  completely  known, 
since  a  and  b  are  given,  and  A  +  B  —  180°—  0.  We  may, 
therefore,  determine  %(A  —  B).  From  this,  and  the 
value  of  |(J.  +  ^0,  the  values  of  A  and  B  can  be  obtained 
by  addition  and  subtraction.  The  remaining  sides  may 
then  be  found  by  the  aid  of  the  law  of  the  sines. 

EXERCISE  XXVIII 

Find  the  remaining  parts  of  the  triangle,  if 
1.   a  =  601,  6  =  289,  and  (7=100°  19'  6". 
SOLUTION.  —  Apply  the  law  of  the  tangents, 

tan  i  (A-E)  =  ^^  tan  }  (A  +  B), 

a+J) 

in  which  a  +  b  =  890,  a  -  b  =  312, 


and         $(A  +  B)  =  £  (180°  -  100°  19'  6")  =  39°  50'  27". 

log  (a  -b)  =    2.49415 

log  tan  \  (A  +  B)  =  9.92136 

12.41551 

log  (a  -f  6)  =  2.94939 
log  tan  I  (A  -  B)  =  9.46612 


CH.  VI,  §  48]  OBLIQUE   TRIANGLES  99 

%(A  -  B)  =  16°  18'  14" 

But  i.  (.4  +  1?)  =  39°  50'  27" 

Hence  A  =  56°    8'  41", 

and  B  =  23°  32'  13". 

The  remaining  side  may  now  be  found  by  the  law  of  the 
sines,  as  in  Case  I. 


.  a  =  23.34,   6  =  55.72,   (7=  18°  23'. 

3.  a  =  576,    6  =  431,   C=  73°  16'  10". 

4.  c  =  .523,   a  =.726,  5  =  50°  28". 

5.  5  =  .0073,   c  =  .008,  .4  =  100°. 
^6.  a  =  54.734,   c  =  65.791,  5  =  105°  54'. 

*  7.  a  =  1673,   5  =  2432,   (7=  98°  5'  15". 

•  8.  a  =  3184,   6  =  917,   C=34°9'16". 

9.  a  =  31.84,    6  =  12.34,   C  =  88°  12'  40". 

10.  a  =  14.59,   6  =  3.99,   C  =  92°  11'  18". 

48.  CASE  III.  Given  two  sides  and  the  angle  opposite 
one  of  them.  —  Let  the  given  parts  be  A,  a,  and  b. 

It  is  shown  in  plane  geometry  that,  when  two  sides 
of  a  triangle  and  the  angle  opposite  one  of  them  are 
given,  there  may  be  constructed,  sometimes  two  triangles, 
sometimes  one,  and  sometimes  no  triangle,  according 
to  the  relation  existing  between  the  given  sides  and 
angles. 

If  the  given  angle  A  is  acute,  and  the  opposite  side  a  is 
less  than  the  adjacent  side  b  but  greater  than  the  perpen- 
dicular CD  (=  b  sin  A),  there  are  two  possible  construc- 
tions. Fig.  56  (a).  ^^ 


100 


PLANE  T1UGONOMETRY 


[On.  VI,  §  48 


If  A  is  acute,  and  a  =  b  sin  A  or  a  >  5,  there  is  one  con- 
struction.    Fig.  56  (£). 


O    o 


(6)  (0 

FIG.  56. 


If  A  is  acute,  and  a<b  sin  A,  no  construction  is  possible. 
Fig.  56  O). 

If  A  is  obtuse,  there  will  be  one  construction  if  a  >  5; 
otherwise  tnere  will  be  no  construction.  Fig.  56  (d). 

Many  of  these  results  appear  also  in  the  trigonometric 
solution  of  the  triangle.  From  the  law  of  the  sines, 


sin  B  = 


b  sin  A 


Since  sin  B  cannot  be  greater  than  1,  we  see  at  once 
that  there  will  be  no  solution  if  a  <  b  sin  A.  Also  that 
there  will  be  only  one  solution  (.#=90°)  when  a  =  bsmA. 

When  a  >  b  sin  A,  there  will  be  apparently  two  values 
of  B,  one  acute  and  the  other  obtuse,  which  are  supple- 
mentary. But  both  of  these  apparent  values  are  not 
always  possible  ;  for,  if  a  >  6,  plane  geometry  tells  us, that 
A  >  B,  and  hence  B  cannot  be  obtuse. 

Again,  if  A  is  obtuse,  B  must  be  acute,  and  there  can  be 
only  one  solution.  Bat  here  there  will  be  no  solution 
when  a<b-,  for,  if  a  <  6,  A  <  B,  and  this  is  impossible 
when  A  is  obtuse. 


CH.  VI,  §  48]  OBLIQUE   TRIANGLES  101 

This  discussion  may  be  condensed  into  the  following 
table  : 

When  A  <  90°,  and  b  sin  A  <  a  <  b,  there  are  two  solutions. 

When  A  <  90°,  and  a^b,  there  is  one  solution. 

When  A  <  90°,  and  a  =  b  sin  A,  there  is  one  solution. 

When  A  <  90°,  and  a<.bsiuA,  there  is  no  solution. 

When  A  >  90°,  and  a>b,  there  is  one  solution. 

When  A  >  90°,  and  a<b,  there  is  no  solution. 

Before  beginning  the  solution  of  any  problem,  the 
student  should  determine  the  number  of  possible  solu- 
tions. If  there  is  one  solution,  the  law  of  the  sines 

gives  the  value  of  B.     Then  (7=180° 


—  (A  +  B).     The  side  c  may  be  found  from  the  formula 

_  a  sin  C 
sin  A 

When  there  are  two  solutions,  the  method  of  procedure 
is  the  same,  except  that  there  will  be  a  second  value  of  B, 
which  we  shall  call  B'  (=  180  -  B).  Then 


It  is  always  possible  to  determine  the  number  of  solu- 
tions by  attempting  to  solve  the  triangle.  If  there  is  no 
solution,  we  shall  obtain  a  positive  value  of  log  sin  B, 
which  is  impossible,  since  sin  B  cannot  be  greater  than 
unity.  Again,  if  we  attempt  to  obtain  two  solutions 
where  is  only  one,  we  shall  find  that  the  sum  of  two  of 
the  angles  of  the  triangle  is  greater  than  180°. 


102  PLANE  TRIGONOMETRY  [On.  VI,  §  18 

EXERCISE  XXIX 

Find  the  remaining  parts  of  the  triangle,  if 
I.    a  =  6,  6=8,  and  .4  =  40°. 

SOLUTION.  —  Since  A  is  acute  and  b  >  a  >  6  sin  A,  there 
two  solutions. 

sin  B  =  —  ,  or  log  sin  B  =  log  b  -f  log  sin  A  —  log  a. 

ci 

log  b=    0.90309 

log  sin  A=   9.80807 

10.71116 

log  a  =  0.77815 
log  sin  B=  9.93301 

Hence    B=    58°  59' 15", 
and  £'  =  121°    0'45". 

C=  180° -(A  +  B)  =  81°   0'45", 

C"  =  180°  -  (A  +  B1)  =  18°  59'  15". 

_  a  sin  (7  f  _  a  sin  (7' 

sin  yl  sin  A 

loga=   0.77815  loga=   0.77815 

log  sin  C  =  J&99464  log  sin  (7 '  =    9.51236 

1O77279  10.29051 

log  sin  A  =  _9.80807  log  sin  A  =   9.80807 

logc=   0.96472  logc'=   0.48244 

c=   9.2198.  c'?=   3.037. 

2.  a  =  77.04,  6  =  91.06,  5  =  51°9r6". 

3.  a  =  80,  6  =  401,  5  =  84°  16' 31". 

4.  a  =  319,  c  =  481,  A  =  41°  32'  40;;. 

5.  a  =  695,  6  =  345,  B  =  21°  14'  25". 


CH.  VI,  §  49]  OBLIQUE   TRIANGLES  103 


6.  a  =  4.32,  6  =  7.61,  B  =  59°  14'. 

7.  a  =  704,  6  =  302,  B  =  25°  14'  13". 

8.  a  =  49,  6  =  45,  £  =  17°  41' 9". 

.  a  =  242,  6  =  767,  E  =  36°  53'  2". 

10.  c  =  1042,  6  =  55.8,  £  =  32°  22'  42". 


49.    CASE  IV.      Given  the  three  sides.  —  If  we  solve 
the  equation  a2  =  62  -f-  c2  —  2  &GJ  cos  ^4.  for  cos  -4, 


^    _ 

we  nave  cos  A  =  —          -  . 

2  be 

The  corresponding  formulas  for  the  other  angles  are 


C          «    _ 

cos  B  =  —  —  -  ,  and  cos  C  = 


,  . 

2  ca  2  ab 

When  the  sides  of  a  triangle  are  expressed  by  small 
numbers,  the  angles  may  be  found  easily  from  these 
formulas,  with  the  aid  of  a  table  of  cosines.  Each  angle 
should  be  found  in  this  way,  and  the  accuracy  of  the 
result  tested  by  adding  the  three  angles. 

EXERCISE  XXX 

1.    Find  the  three  angles,  if  a  =  8,  6  =  5,  c  =  7. 

SOLUTION.        cos  A  =  25  +  49  ~  64  =  .1429.       A  =  81°  47'. 

70 

cos  B  =  64  +  49f~25  =  .7857.       B  =  38°  13'. 
11.Z 

0=60°. 


80 

Adding,  A  +  B  4-  C  =  180°,  and  the  solution  is  accurate  as 
far  as  minutes  ;  but  since  we  are  using  a  four-place  table,  we 
cannot  determine  the  seconds.  With  a  five-place  table,  the 
seconds  may  be  determined  with  fair  accuracy  ;  but  even  with 


104 


PLANE  TRIGONOMETRY 


|Cn.  VI,  §  50 


a  five-place  table  the   results  may  often  differ  a  number  of 
seconds  from  the  true  value. 
Find  the  three  angles,  if 


2.  a  =  13,  b  =  8,  c  =  15. 

3.  a  =  26,  b  =  31,  c.=  21. 


4.  a  =  25,  6  =  26,  c  =  27. 

5.  a  =  17,  6  =  20,  c  =  27. 


50.  Solution  by  logarithms.  —  When  the  sides  of  the 
triangle  are  expressed  by  large  numbers,  this  method  is 
long,  since  the  formulas  are  not  adapted  to  the  use  of 
logarithms.  They  may,  however,  be  changed  into  a  dif- 
ferent form,  to  which  logarithms  may  be  applied. 

From  formula  [23], 


sin     A  = 


1  —  cos  A. 


Substituting  in  this  the  value  of  cos  A  obtained  in  Art.  49, 


1- 


sin     A  = 


2  be 


J 
* 


±be 


O-  ft  +  c)(a+b-c) 
4  be 


For  convenience,  let  a  +  b  +  c  =  2  s.     Then  a  —  b  +  6 
2(*  -  5),  and  a  -f  b  -  c  =  2(s  —  c). 
Substituting  these  values, 


CH.  VI,  §  50]  OBLIQUE   TRIANGLES  105 

In  like  manner, 


2JC 
cos  £  A  = 


Either  of  these  formulas  might  be  used  for  solving  tri- 
angles, when  the  three  sides  are  given,  arid  are  well 
adapted  to  the  use  of  logarithms.  But  for  values  of  ^  A 

near    — ,  the  first  is  inaccurate   (since   the  sine  of  such 

angles  changes  very  slowly)  and  the  same  is  true  of  the 
second  for  very  small  values  of  |-  A.  It  is,  therefore,  best 
to  obtain  the  formula  for  tan  J  A,  which  may  be  used  for 
all  angles, 


8  —  a 


If  we  let        J(* -*X*- 

s 

this  formula  becomes 


[85,  a] 


106  PLANE  TRIGONOMETRY  [Cii.  VI,  §  50 

Since  r  is  not  changed  by  interchanging  the  letters,  the 
corresponding  formulas  for  the  other  angles  are 


^\C  =  ^.  [35,  e] 

These  formulas  are  evidently  well  adapted  to  the  use  of 
logarithms,  and,  since  the  tangent  varies  rapidly  for  angles 
of  any  magnitude,  they  may  be  used  in  all  cases.  Each 
of  the  angles  should  be  found  by  these  formulas,  and  the 
accuracy  of  the  work  tested  by  adding  the  three  values,, 
With  a  five-place  table  the  sum  should  not  differ  by  more 
than  a  few  seconds  from  180°. 

EXERCISE  XXXI 

1.    Find  the  angles  of  the  triangle,  if  a  =  15.47,  b  =  17.39, 

c  =  22.88. 

SOLUTION. 


.   a  =  15.47 
b  =  17.39 
c  =  22.88 

log  (s  - 
log  (s  - 
log  (s  - 

a)  =  1.09342 
b)  =  1.02036 
c)  =  0.69810 

2  s  =  55.74 
s  =  27.87 
s  -  a  -  12.40 
s  -  b  =  10.48 
s—c=  4.99 

logs 
log  r- 

2.81188 
=  1.44514 
=  1.36674 

log  r 

=  0.68337 

Subtracting  log  (s  —a),  log  (s  —  b),  and  log  (s  —  c)  successively 

from  log  r,  we  have  log  tan  £  A  =  9.58995,  or  A  =  42°  30'  44", 

log  tan  i  B  =  9.66301,  or  B  =  49°  25'  49", 

log  tan  i  C  =  9.98527,  or  C  =  88°    3' 27". 

180°  00  00. 


CH.  VI,  §  51]  OBLIQUE   TRIANGLES  107 

Find  the  three  angles,  if 

2.  a  =  17,  b  =  113,  c  =  120. 

3.  a  =  3359.4,  b  =  4216.3,  c  =  4098.7. 

4.  d  -3.9009,  6  =  2.7147,  c  =  3.0012. 
^X  5.    a  =  289,  6  =  601,  c  =  712. 

6.  a  =  354.4,  6  =  277.9,  c  =  401.3. 

7.  a  =  5.134,  6  =  7.268,  c  =  9.313. 

8.  a  =  0.099,  6  =  0.101,  c  =  0.158. 

9.  a  =  33.112,  6  =  44.224,  c  =  55.336. 

51.    Area  of  an  oblique  triangle.  —  Let  K  denote  the  area 
of  the  triangle  ABC.     Draw  CD  perpendicular  to  AB. 

G 

h 
D 


e       D 
FIG.  57  (a). 


c         B 
FIG.  57  (b). 


Then  K 

But  in  either  figure     CD  =  a  sin  B. 


Hence 


=  ~  ac  sin  B. 


[36] 


In  like  manner  it  may  be  shown  that 

K=  |  ab  sin  C  =  J  be  sin  A. 

Or,  the  area  of  any  triangle  is  equal  to  one-half  of  the 
product  of  any  two  sides  and  the  sine  of  the  included  angle. 


108  PLANE  TRIGONOMETRY  [CH.  VI,  §  51 

When  the  three  sides  of  a  triangle  are  given,  the  area 
may  be  obtained  as  follows  : 

By  formula  [20]     sin  B=Z  sin  J-  B  cos  |  B. 
Hence  K—  ac  sin  |  B  cos  ^B. 


Substituting  in  this  expression  the  values  of  sin  |  B  and 
cos     B  obtained  in  Art.  50,  we  have 


K=  --  «  «- 


. 

* 


ac 

K  =  Vs(s-a)(s-  &)(s-c).  [37] 

When  any  other  three  parts  are  given,  find  either  two 
sides  and  the  included  angle  or  the  three  sides,  and  apply 
one  of  the  above  formulas. 

EXERCISE  XXXII 
Find  the  area  of  each  of  the  following  triangles  : 

1.  a  =  5,  6  =  6,  (7=78°  9'. 

2.  a  =  45.34,  c  =  56.45,  B  =  100°  10'. 

3.  6  =  .1001,  c  =  .3204,  A  =  30°  33'  25". 

4.  a  =  14,  6  =  14,  c  =  15. 
^   5.    a  =.39,  b  =  .8,  c=.89. 

^6.    a  =  56,  b  =  90,  c  =  106. 

v7.    a  =  318,  6  =  181,  ^1  =  64°  58'. 

8.6  =  34.51,  c  =  183.94,  A  =  23°  53'  17". 
<  9.    a  =  .7845,  6  =  .07859,  (7=  120°  43'  50". 

10.    a  =  23,  A  =  76°  53'  25",  B  =  13°  29'  15". 


CH.  VI,  §  51]  OBLIQUE   TRIANGLES  109 

EXERCISE  XXXIII. 

/-I.  The  diagonals  of  a  parallelogram  are  73  and  95,  and  they 
cross  each  other  at  an  angle  of  35°  28'.  Find  the  sides  and 
angles  of  the  parallelogram. 

2.  At  one  point  of  observation  the  horizontal  angle  sub- 
tended by  a  round  fort  is  4°  35'.  On  going  500  ft.  directly 
toward  the  fort,  it  is  found  to  subtend  an  angle  of  6°.  Find 
the  diameter  of  the  fort. 

^-  3.  The  parallel  sides  of  a  trapezoid  are  16  and  23ft.  The 
angles  at  the  extremities  of  the  longer  side  are  35°  54'  and 
76°  20 '.  Find  the  non-parallel  sides. 

„  4.  A  tower  stands  at  the  foot  of  an  inclined  plane  whose 
inclination  to  the  horizon  is  9°;  a  line  100  ft.  in  length  is 
measured  straight  up  the  inclined  plane  from  the  foot  of  the 
tower,  and  at  the  upper  extremity  of  this  line  the  tower  sub- 
tends an  angle  of  54°.  Find  the  height  of  the  tower. 

5.  Looking  out  of  a  window,  with  his  eye  at  the  height  of 
15  ft.  above  the  roadway,  an  observer  finds  that  the  angle  of 
elevation  of  the  top  of  a  telegraph  pole  is  17°  18' 35",  and  the 
angle  of  depression  of  its  foot  is  8°  32'  15".     Find  the  height 
of  the  pole  and  its  distance  from  the  observer. 

6.  A  and  B  are  two  points,  200  yards  apart,  on  the  bank  of 
a  river,  and  C  is  a  point  on  the  opposite  bank.     The  angles 
ABC  and  BAC  are  respectively  54°  30'  and  65°  30'.     Find  the 
breadth  of  the  river. 

7.  A  ship  sails  due  east  past  two  headlands  which  are  two 
miles  apart  and  bear  in  a  line  due  south.     Half  an  hour  later 
one  of  the  headlands  bears  15°  south  of  west  and  the  other  30°. 
What  is  the  rate  of  the  ship  ? 

8.  What  is  the  approximate  distance  at  which  a  boy  must 
hold  a  coin  one  inch  in  diameter  from  his  eye  to  conceal  the 
moon,  if  its  apparent  angular  diameter  is  half  a  degree  ? 


110  PLANE  TRIGONOMETRY  [Cn.  VI,  §  51 

9.  A  balloon  rises  vertically  at  a  horizontal  distance  of 
3000  yards  from  an  observer,  who  finds  the  angle  of  elevation 
to  be  15°  when  he  first  sights  the  balloon.  When  he  again 
measures  the  angle  he  finds  it  to  be  30°.  Through  what  dis- 
tance has  the  balloon  risen  between  the  two  observations  ? 

10.  A  pole  10  ft.  high  stands  upright  in  the  ground.     The 
angle  of  elevation  of  the  top  of  a  tree  from  the  foot  of  the 
pole  is  32°  27',  while  the  angle  of  elevation  of  the  top  of  the  pole 
from  the  foot  of  the  tree  is  22°  44'.     Find  the  distance  between 
the  pole  and  the  tree,  also  the  height  of  the  tree. 

11.  A  telegraph  pole  stands  on  the  bank  of  a  stream.     Its 
angle  of  elevation  from  a  point  directly  opposite  on  the  other 
bank  is  36°  53',  and  from  a  point  60  ft.  from  the  bank  in  a 
straight  line  with  the  first  point  and  the  pole,  the  angle  is 
16°  42'.     Find  the  width  of  the  stream  and  the  height  of  the 
telegraph  pole. 

12.  A  church  stands  on  the  bank  of  a  river.     From  the 
opposite  side  of  the  river  the  angle  of  elevation  of  the  top  of 
the  spire  is  found  to  be  57°  25'.     The  observer  moves  back 
200  ft.  in  a  direct  line  with  the  foot  of  the  spire  and  there 
finds  the  angle  of  elevation  to  be  48°  30'.     Find  the  width  of 
the  river. 

13.  A  man  wishing  to  determine  roughly  the  length  of  a 
pond  finds  that  a  line  joining  two  stakes,  driven  one  at  each 
end  of  the  pond,  runs  N.W.     He  then  takes  150  paces  from 
one  of  the  stakes  toward  the  N.E.,  turns  and  takes  200  paces 
to  the  other  stake.     What  is  the  length  of  the  pond,  if  one  of 
the  man's  paces  is  2-J-  ft.  ?     What  is  the  angle  through  which 
the  man  turns  ? 

14.  The  angle  of  elevation  of  a  steeple  is  71°  34',  when  the 
observer's  eye  is  on  a  level  with  the  bottom.     From  a  window 
25  ft.  above  the  place  where  the  observer  stands  the  angle  of 
elevation  is  69°  27'.     Find  the  observer's  distance  from  the 
steeple  and  the  height  of  the  steeple. 


Cn.  VI,  §  51]  OBLIQUE   TRIANGLES  111 

15.  A  man  at  a  station  B  at  the  foot  of  a  mountain  observes 
the  elevation  of  the  summit  A  to  be  50°.     He  then  walks  one 
mile  toward  the  summit  up  an  incline,  making  an  angle  of  30° 
with  the  horizon  to  a  point  C,  and  observes  the  angle  ACB  to 
be  150°.     Find  the  height  of  the  mountain. 

16.  A  tower  I  ft.  high  stands  in  a  plane.     The  angles  of 
depression  from  the  top  of  the  tower  of  two  objects  lying  in  the 
plane  in  a  direct  line  from  the  foot  of  the  tower  are  for  the 
nearer  a  and  for  the  more  remote  /3.     Find  the  distance  between 
the  objects. 

17.  Two  inaccessible  objects  P  and  Q  lie  in  a  horizontal 
plane.     To  find  the  distance  PQ  a  base  line  AB  of  500  yards  is 
measured  in  the  plane.     At  its  extremities  A  and  B,  the  follow- 
ing angles  are  measured:  ZBAQ  =  36°12',  ZQAP=50°46', 
Z  ABP  =  43°  22',  and  PBQ  =  72°  9'.     What  is  the  distance  PQ  ? 

18.  There  is  a  tower  on  the  top  of  a  hill.     From  a  point  in 
the  plane  on  which  the  hill  stands  the  angle  of  elevation  of  the 
base  of  the  tower  is  37°,  and  of  the  top  of  the  tower  50°.     From 
another  point  straight  away  from  the  hill  in  a  line  through  the 
first  point  and  200  ft.  from  that  point  the  angle  of  elevation 
of  the  top  of  the  tower  is  31°  22'.    Find  the  height  of  the  tower. 

.  An  obelisk  stands  on  a  hill  whose  slope  is  uniform.  A 
man  measured  from  the  foot  of  the  obelisk  a  distance  of  32  ft. 
directly  down  the  hill  and  found  the  angle  between  the  incline 
and  the  top  of  the  obelisk  to  be  45° ;  after  measuring  down- 
ward an  additional  distance  of  68  ft.,  the  angle  found  in  the 
same  manner  was  21°47f.  What  is  the  height  of  the  obelisk 
and  the  inclination  of  the  hillside  ? 

20.  An  observer  in  a  ship  sees  two  rocks,  A  and  B,  in  the 
same  straight  line  N.  25°  E.     He  then  sails  northwest  for  4 
miles,  and  observes  A  to  bear  due  east  and  B  northeast  of  his 
new  position.     Find  the  distance  from  A  to  B. 

21.  The  circular  basin  of  a  fountain  subtends  an  angle  of 
25°  at  a  distance  of  44  ft.  from  the  edge  of  the  basin,  measured 
on  a  diameter  produced.     Find  the  radius  of  the  basin.  ;^  ,  /  V 


112  PLANE  TRIGONOMETRY  [Cn.  VI,  §  51 

22.  From  the  top  of  a  vertical  tower  whose  height  is  100  ft., 
the  angle  of   depression  of  an  object  is   observed  to  be  60°, 
and  from  the  base  to  be  30°;  show  that  the  vertical  height  of 
the  base  of  the  tower  above  the  object  is  50  ft. 

23.  A  flagstaff  40  ft.  tall  stands  on  a  castle  wall.     At  a 
horizontal  distance  of  20  ft.  from  the  foot  of  the  wall  the 
flagstaff  subtends  an  angle  of  15°.     Find  the  height  of  the 
wall. 

24.  The  angle  of  elevation  of  a  tower  at  a  distance  of  20 
yards  from  its  foot  is  three  times  as  great  as  the  angle  of 
elevation  100  yards   from  the  same  point.      Show  that  the 

height  of  the  tower  is  ~~=  ft. 

25.  Two  parallel  chords  of  a  circle,  lying  on  the  same  side 
of  the  centre,  subtend  respectively  72°  and  144°  at  the  centre. 
Show  that  the  distance  between  the  chords  is  half  the  radius 
of  the  circle. 

26.  A  person  standing  due  south  of  a  lighthouse  observes 
that  his  shadow,  cast  by  the  light  at  the  top,  is  24  ft.  long ;  on 
walking  100  yards  due  east,  he  finds  his  shadow  to  be  30  ft. 
Supposing  him  to  be  6  ft.  high,  find  the  height  of  the  light 
from  the  ground. 

27.  A  man  5  ft.  tall  stands  on  the  edge  of  a  pond.     The 
angle  of  elevation' of  a  tree  on  the  opposite  bank  is  45°  and  the 
angle  of  depression  of  its  reflection  is  60°.     Find  the  height  of 
the  tree. 

NOTE.  —  The  reflection  of  the  top  of  the  tree  appears  as  far  below 
the  surface  of  the  water  as  the  top  of  the  tree  is  above  the  water. 

28.  One  end  of  a  pole  rests  on  the  ground  and  the  other  end 
touches  the  top  of  a  window.     When  the  lower  end  of  the  pole 
is  moved  away  16  ft.  farther  from  the  wall,  the  top  rests  on 
the  sill  of  the  window.     If  the  first  angle  the  pole  makes  with 
the  ground  is  71°  25'  and  the  second  48°  35',  find  the  length  of 
the  window. 


CH.  VI,  §  51]  OBLIQUE   TRIANGLES  113 

29.  Two  captive  balloons  are  floating  at  equal  heights  in 
calm  air.     A  man  standing  in  the  straight  line  between  their 
points  of  attachment  finds  the  angle  of  elevation  of  the  nearer 
balloon  to  be  tan"1  1.     He  then  walks  a  distance  of  240  ft.  at 
right  angles  to  the  straight  line  joining  the  points  of  attach- 
ment and  finds  the  angle  of  elevation  of  the  same  balloon  to  be 
tan"1  -|,  and  that  of  the  other  tan"1  -/-$.     Find  the  height  of  the 
balloons  and  the  distance  between  their  centres. 

30.  Wishing  to  find  the  inclination  of  a  roadway  rising  from 
a  level  park,  a  man  walked  100  ft.  up  the  incline  and  observed 
the  angle  of  depression  of  an  object  in  the  park  to  be  30°.    After 
walking  up  the  plane  100  ft.  farther,  the  angle  of  depression 
of  the  same  object  was  45°.    Show  that  the  angle  of  inclination 
is  cot-1(2—  - 


31.  Two  persons  stand  facing  each  other  on  opposite  sides 
of  a  pool.     They  are  of  such  heights  that  the  eye  of  one  is 
5  ft.  above  the  ground  and  that  of  the  other  6  ft.     When  the 
line  of  vision  of  each  makes  an  angle  of  54°  with  vertical,  the 
reflection  of  the  eye  of  either  is  visible  to  the  eye  of  the  other. 
What  is  the  width  of  the  pool  ? 

32.  Viewed  from  the  S.E.  corner  of  a  room  the  N.E.  corner 
has  an  elevation  of  0  and  the  S.W.  corner  of  <£.     Find  the 
elevation  of  the  N.W.   corner,  also  the  angles  the  diagonals 
make  with  the  edges  of  the  room. 

33.  The  angle  of  elevation  of 
a  tower  from  a  point  A  due  south 
is  «,  and  from  a  point  B  due  west 
of  the  first  station  it  is  ft.     If  the 
distance,    AB,    between    the  two 
stations  is  b,  show  that  the  height 
of  the  tower  is 

6  sin  a  sin 


V  sin  (a  +  ft)  sin  (a  -  ft)  FlG.  58. 


114  PLANE   TRIGONOMETRY  [Cn.  VI,  §  51 

34.  A  man  walked  up  an  inclined  plane  a  feet  and  observed 
the  angle  of  depression  of  an  object  in  a  horizontal  plane  to  be  a. 
When  he  had  walked  a  distance  of  2  a  ft.  further  up  the 
incline  the  angle  of  depression  was  2  a.  Show  that  the  angle 
the  incline  makes  with  the  horizontal  plane  is 


1  —  4  sin-  a 

35.  A  man  walks  along  the  bank,  AB,  of  a  straight  stream 

and  at  A  observes  the  greatest 
angle  subtended  by  two  objects, 
P  and  Q,  on  the  opposite  side 
to  be  75°  ;  he  then  walks  a  dis- 
"tance  of  300  ft.  to  B  and  finds 
that  the  objects  are  in  a  straight 

line,  which  makes  an  angle  of  15°  with  the  bank.     Find  the 

distance  between  the  objects. 

NOTE.  —  The  point  on  the  bank  where  the  greatest  angle,  subtended 
by  the  objects,  is  made,  will  be  the  point  of  tangency  of  a  circle  pass- 
ing through  them.  Z.BPA  =  Z.QAB  =  0  can  be  expressed  in  terms 
of  15°,  75°,  and  90°. 

36.  A  person  on  the  top  of  a  tower  observes  the  angles  of 
depression  of  two  objects  in  the  plane  on  which  the  tower 
stands  to  be  60°  and  30°.     He  knows  the  distance  between  the 
two  objects  to  be  500  ft.     The  angle  subtended  at  his  eye  by 
,the  line  joining  the  two  objects  is  30°.     Find  the  height  of  the 
tower. 

37.  From  two  stations  a  ft.  apart  a  balloon  is  observed. 
At  one  of  the  stations  the  horizontal  angle  between  the  balloon 
and  the  other  station  is  y,  and  the  angle  of  elevation  of  the 
balloon  is  a;  at  the  other  station  the  corresponding  angles  are  8 
and  ft.     Show  that  the  height  of  the  balloon  is 

a  sin  y  tan  (3 
sin  (8  +  y) 

Also  show  that        tan  a  sin  8  =  tan  ft  sin  y. 


CH.  VI,  §  51] 


OBLIQUE   TRIANGLES 


115 


38.    An  inclined  plane  AB,  of  length  I,  has  a  vertical  rod,  PQ, 
fastened  to  its  surface  at  such  a  distance  from  its  foot  that  the 
upper  end  of  the  rod  and  the  top 
of  the  plane  are  in  the  same  straight 
line  with  a  point  D  at  a  distance  I 
from  the  foot  of  the  plane.     The 
angles  subtended  by  the  rod  and    < 
the   part   of  the   plane  below  are 
each  equal  to  a.    Find  the  distance 
at  the  foot  of  the  rod  from  the  lower  end  of  the  plane,  also  the 
length  of  the  rod. 


A 

FIG.  60. 


PART   II 
SPHERICAL   TRiaONOMETRY 

CHAPTER   VII 

RIGHT  AND  QUADRANTAL  TRIANGLES 

52.  A  spherical  triangle  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  the  arcs  of  three  great  circles  which 
intersect.  Spherical  Trigonometry  is  concerned  with  the 
relations  of  the  sides  and  angles  of  a  spherical  triangle, 
and  the  computation  of  the  unknown  parts  when  any 
three  parts  are  given.  In  the  following  treatment  a 
knowledge  of  Solid  Geometry  is  presupposed,  but  it  is 
thought  best  to  begin  with  a  statement  of  the  definitions 
and  theorems  which  are  most  important  for  our  subject. 

Let  ABO  be  a  spherical 
triangle  on  a  sphere  whose 
centre  is  at  0.  Join  0  to 
the  vertices  A,  B,  and  (7, 
and  pass  planes  through  0 
and  the  sides  of  the  triangle. 
These  sides  are  measured 
in  degrees,  and  their  measures 
are,  therefore,  equal  to  the  measures  of  the  correspond- 
ing plane  angles  A  OB,  BOO,  00 A. 

117 


118  SPHERICAL   TRIGONOMETRY         [Cn.  VII,  §  52 

A  spherical  angle  is  equal  to  the  angle  between  tan- 
gents to  the  sides  of  the  angle,  drawn  at  the  vertex. 
Hence  the  angle  A  is  equal  to  the  angle  formed  by  draw- 
ing, in  the  faces  A  OB  and  AOC,  any  two  lines  perpen- 
dicular to  OA  at  the  same  point. 

A  spherical  angle  may  also  be  measured  by  the  arc  of  a 
great  circle  having  the  vertex  of  the  angle  as  a  pole  and 
intercepted  between  its  sides. 

We  shall  restrict  our  study  of  triangles  to  those  in 
which  each  of  the  angles  and  sides  is  less  than  180°.  The 
sum  of  the  three  sides  may  have  any  value  less  than  360°, 
while  the  sum  of  the  angles  must  lie  between  180°  and 
540°. 

A  triangle  may  contain  one,  two,  or  three  right  angles, 
or  each  of  the  angles  may  be  greater  than  90°.  If  each 
of  the  angles  is  a  right  angle,  each  of  the  sides  is  a  quad- 
rant, and  it  is  called  a  tri-rectangular  triangle. 

Polar  triangles  are  so  related  that  the  vertices  of  each 
are  the  poles  of  the  corresponding  sides  of  the  other. 
Each  angle  of  either  triangle  is  the  supplement  of  the 
side  lying  opposite  it  in  its  polar  triangle. 

When  one  or  more  of  the  sides  of  a  spherical  triangle 
are  quadrants,  it  is  called  a  quadrantal  triangle. 

EXERCISE  XXXIV 

1.  Prove  that,  if  a  triangle  has  three  right  angles,  it  is  its 
own  polar. 

2.  Prove  that  the  polar   of  a  right  triangle  is  a  quadrantal 
triangle. 

3.  Prove  that,  if  a  triangle  has  two  right  angles,  the  sides 
of  the  polar  triangle  opposite  these  are  quadrants,  and  that 
the  third  side  measures  the  third  angle  of  the  given  triangle. 


CH.VII,§53]    RIGHT  AND  QUADEANTAL  TRIANGLES  119 

4.  Prove  that,  if  one  side  of  a  triangle   is   a   quadrant, 
either  of  the  other  sides  and  the  angle  opposite  it  are  either 
both  less  or  both  greater  than  90°. 

5.  Prove  that,  if  a  triangle  has  one  right  angle,  each  of  its 
remaining  angles  is  in  the  same  quadrant  as  the  side  oppo- 
site it. 

6.  Prove  that,  if  the  sides  about  the  right  angle  of  a  right 
spherical  triangle  are  in  the  same  quadrant,  the  hypotenuse  is 
less  than  90°;  while  if  they  are  in  different  quadrants  the 
hypotenuse  is  greater  than  90°. 

7.  Prove  that,  if  one  of  the  sides  of  a  right  triangle  is  equal 
to  the  opposite  angle,  the  remaining  parts  are  each  equal  to  90°. 

8.  The  angles  of  a  triangle  are  80°,  75°,  and  105°.     Find 
the  number  of  degrees  in  each  side  of  the  polar  triangle,  and,  if 
the  radius  of  the  sphere  be  90  ft.,  compute  their  lengths  in  feet. 

9.  The  sides  of  a  spherical  triangle  are  70°,  80°,  and  110°. 
Find  the  angles  of  its  polar  triangle. 

10.  In  an  equiangular  spherical  triangle  each  of  the  angles 
is  120°.  Find  the  value  of  each  side  of  the  polar  triangle.  If 
the  angles  increase  to  180°  each,  state  the  limits  of  the  tri- 
angle and  its  polar. 

53.  Right  triangles.  — Let  the  spherical  triangle  ABO 
(Fig.  2)  be  right-angled  at  (7,  and  let  each  of  the  other 
parts  be  less  than  90°.  Pass  planes  through  the  sides 
of  the  triangle  and  0,  the  centre  of  the  sphere.  Repre- 
sent the  measure  of  the  sides  opposite  A,  B,  and  0  by 
the  corresponding  small  letters  a,  5,  c.  Then  these  are 
also  the  measures  of  the  corresponding  angles  at  0. 

That  is, 

Z.BOO=a,   ZCOA  =  b,   ZAOB=c. 

Through  B  pass  a  plane  BED  perpendicular  to  OA. 
Then  EB  and  ED  are  perpendicular  to  OA,  and  the 


120 


SPHERICAL   TRIGONOMETRY          [Cn.  VII,  §  53 


angle  BED   is  the   measure  of   the  spherical   angle  A. 

Also  BD  is  perpendicular  to 
the  plane  AOC,  since  it  is 
the  line  of  intersection  of  two 
planes  which  are  perpendicu- 
lar  to  that  plane.  Then  the 
triangles  BOE,  BOD,  DOE, 
and  BDE  are  all  right  triangles. 


FIG.  2. 


Then 


COS  G  = 


OE 
OB 


Also        OE  =  OD  cos  b  =  OB  cos  a  cos  b. 


Hence 
Again 
Hence 


—          — 


cos  c  =  cos  a  cos  b. 

1  a 


EB      OB  sin  c     sin  c 
sin  a  =  sine  sin  A. 

Interchanging  a  and  b,  A  and  J5, 


sin  6  =  sin  c  sin  J5. 


Again 


cos  J. 


tan 


0^  tan  c      tan 
or  cos  A  =  tan  6  cot  c. 

Also  cos  .B  =  tan  a  cot  c. 

Formula  [4]  may  be  written 

sin  b  cos  c 


cos  A  = 


But 


sin 


cos  5  sin  c 
=  sin  .#,  and 


[2] 
[3] 


[5] 


cos  6 


=  cos  a. 


CH.VII,§53]    RIGHT  AND   QUADRANTAL  TRIANGLES          121 

Hence  cos  A  =  cos  a  sin  B.  [6] 

Also  cos  B  =  cos  b  sin  A.  [7] 

Substituting  the  values  of  cos  a  and  cos  b  obtained  from 
these  last  equations  in  formula  [1],  we  have 

cos  c  =  cot  A  cot  B.  [8] 

A     .  .    ,      ED     DBcotA     cc&  A 

Again  sin  b  =  — —  =  —  —  =  —    — • 

OD      DB  cot  a     cot  a 

Hence  sin  b  =  tan  a  cot  A.  [9] 

Also  sin  a  =  tan  b  cot  B.  [10] 

In  deriving  these  formulas  we  have  used  a  triangle  in 
which  none  of  the  parts  is  greater  than  90° ;  but  they 
may  be  easily  shown  to  hold 
for  any  right  triangle.  Let 
ABO  (Fig.  3)  be  a  right  tri- 

A. 

angle  in  which  a  <  90°  and 
I  >  90°.  Then  by  Problem  6, 
Exercise  xxxiv,  c> 90°.  Con- 
tinue the  sides  AB  and  A  O  until  they  meet  at  Af. 
A' BO  is  a  right  triangle  in  which  each  of  the  five  parts 
is  less  than  90°. 

Since  each  of  the  arcs  ABA'  and  AC  A'  is  a  semicir- 
cumference, 

A'B  =  180°  -  <?,  and  OAr  =  180°  -  b. 

Then  by  formula  [1], 

cos  (180°  -  c)  =  cos  a  cos  (180°  -  5), 
or  cos  c=  cos  a  cos  b. 

Let  the  student  show  that  all  the  other  formulas  hold 
for  this  case.     Let  him  also  construct  a  figure,  and  prove 


122  SPHERICAL   TRIGONOMETRY          [Cii.  VII,  §  54 

that  these  formulas  hold  when  both  a  and  b  are  greater 
than  90°. 

54.    Napier's  rules  of  circular  parts. — As  it  is  difficult 
to   remember  these   ten    formulas,   the  following   device 
co  B        may  be   used   as   an   aid   to  the 
memory.     The  two   legs   of  the 
right   triangle,  the  complements 
^  of  the  two  angles,  and  the  com- 

plement    of   the   hypotenuse    are 
FlG-  *•  called   the  circular  parts.     Place 

these  on  the  triangle  as  shown  in  Fig.  4,  omitting 
the  right  angle.  Any  one  of  these  five  parts  may 
be  called  the  middle  part;  then  the  two  parts  on  each 
side  of  it  are  called  the  adjacent  parts  and  the  re- 
maining two,  the  opposite  parts.  Then  the  ten  formulas 
obtained  above  may  be  condensed  into  the  two  following 
rules : 

The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
tangents  of  the  adjacent  parts. 

The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
cosines  of  the  opposite  parts. 

If  we  apply  these  rules,  using  each  part  successively 
as  the  middle  part,  they  will  be  found  to  give  the  ten 
formulas  of  the  previous  article. 

For  example,  if  a  is  taken  as  the  middle  part,  b  and 
co.  B  are  the  adjacent  parts,  and  co.  A  and  co.  c  are  the 
opposite  parts.  Then  the  first  rule  gives  sin  a  =  tan  b 
tan  co.  B  =  tan  b  cot  B,  which  is  formula  [10].  The 
second  rule  gives  sin  a  =  cos  co.  A  cos  co.  c  =  sin  A  sin  <?, 
which  is  formula  [2]. 


CH.VH,  §55]    RIGHT  AND  QUADRANTAL  TRIANGLES  123 

55.  Solution  of  right  triangles.  —  When  any  two  of  the 
parts  of  a  right  spherical  triangle  are  given,  the  remaining 
parts  may  be  determined  by  the  aid  of  the  ten  formulas 
of  Art.  53.  Six  cases  may  occur.  There  ma}r  be  given 

1.  the  two  legs, 

2.  one  leg  and  the  hypotenuse, 

3.  the  two  angles, 

4.  one  angle  and  the  adjacent  side, 

5.  one  angle  and  the  opposite  side, 

6.  one  angle  and  the  hypotenuse. 

There  will  be   one   determinate  solution  in  all  of  these 
cases  except  case  5,  where  there  may  be  two  triangles 
which  satisfy  the  given   conditions.     This  appears  geo- 
metrically  if    the    sides   AC 
and  AB  of  the  triangle  ABO 
are  extended  to  form  a  lune. 
The  angle  A'  equals  angle  A, 
and  the  triangles  ABO  and 
A'BC    are     right     triangles 
which  contain  a  given  angle  A  and  the  opposite  side  a. 

The  formula  required  for  the  solution  of  any  problem 
is  best  obtained  by  marking  the  two  given  parts  and  the 
one  required  on  Fig.  4.  Then  choose  for  the  middle  part 
that  one  of  the  three  which  has  the  other  two  either  as 
adjacent  or  opposite  parts.  For  example,  if  a  and  B  are 
given  and  c  is  wanted,  co.  B  should  be  chosen  as  the 
middle  part.  Then  sin  co.  B  =  tan  co.  c  tan  a,  or  cos  B 
=  tan  a  cot  c,  from  which  c  may  be  obtained. 

If  a  and  c  are  given  and  A  is  wanted,  a  must  be  chosen  as 
the  middle  part ;  then  co.  A  and  co.  c  are  the  opposite  parts. 


124 


SPHERICAL   TRIGONOMETRY          [Cii.  VII,  §  55 


When  each  of  the  parts  of  the  spherical  triangle  which 
are  given  is  less  than  90°,   the  solution   by  the  aid  of 

logarithms  presents  no  new 
difficulty.  But  if  either  of  the 
parts  is  greater  than  90°,  care- 
ful  attention  must  be  paid  to 
the  signs  of  the  functions.  The 
following  example  illustrates 
the  method  of  procedure. 

EXAMPLE  1.    Find  the  remaining  parts  of  the  right 
spherical  triangle   in  which  a  =  125°  and  b  =  60°. 
By  Napier's  rules  we  find  that 


cos  G  =  cos  a  cos 


and 


sin  b  =  tan  a  cot  A,  or  tan  A  = 
sin  a  =  tan  b  cot  B,  or  tan  B  = 


tan  a 
sin  b 

tan  5 
sin  a 


Since  a  is  between  90°  and  180°,  cos  a  and  tan  a  are  nega- 
tive, while  sin  a  is  positive.  Then  c  and  A  are  obtuse, 
and  B  is  acute. 


log  cosa  =  9.75859 
log  cos  b  =  9. 69897 
log  cos  c  =  9.45756 

180°-  c=    73°  20'    3", 
c  =  106°  39'  57". 


log  tan  a  =  10. 15477 
log  sin  b  =  9.93753 
log  tan  A  =  10.21724 

180°-  A  =    58°  46'    2", 
A  =  121°  13'  58". 


log  tan  5  =  10.23856 
log  sin  a  =  9.91336 
log  tan  B=  10.32520 

.g=64°41'  20". 


CH.  VJI,  §  55]    RIGHT  AND  QUADRANTAL  TRIANGLES  125 

EXAMPLE  2.    Find  the   remaining  parts  of  the   right 

spherical  triangle  in  which  a  =  21°  39'  and  J.  =  42°  10'  10". 

The  formulas  needed  for  the  solution  in  this  case  are 

found  to  be  sin  c  =  —    — ,  sin  b  =  tan  a  cot  A,  sin  B  = . 

sin  A  *  os  a 

Since  each  of  the  unknown  parts  is  to  be  determined  by 
its  sine,  there  will  be  two  values  of  each  less  than  180°. 
From  Example  5,  Exercise  xxxiv,  it  appears  that  a  and 
A  must  be  in  the  same  quadrant  ;  it  is  also  evident  from 
the  above  formulas  that  there  will  be  no  solution  unless 
sin  A  >  sin  a. 

log  sin  a  =  9.56695  log  tan  a  =    9.59872 

log  s'mA  =  9.82693  log  cot 4  =  10.04298 

log  sin  c  =  9.74002  log  sin  b  =    9.64170 

c  =    33°  20'  15",  b  =    25°  59'  28", 

or         =  146°  39'  45".  or         =  154°    0'  32". 

log  cos^4=  9.86991 
log  cos  a  =  9.96823 
log  sin  5=  9.90168 

S  =    52°  53', 
or         =  127°    1'. 

These  values  must  be  combined  according  to  the  laws 
stated  in  Examples  5  and  6  in  Exercise  xxxiv.  In  this 
problem  all  the  values  less  than  90°  form  one  solution 
and  those  greater  than  90°,  the  other.  This  will  be  true 
only  in  case  the  given  parts  are  less  than  90°. 


126  SPHERICAL   TRIGONOMETRY          [Cti.  VII,  §  56 

EXERCISE  XXXV 

Find  the  remaining  parts  of  the  right  spherical  triangle  in 
wnich 

1.  a  =  9°  45'  19",  b  =  12°  16'  42". 

2.  a  =  28°  26'  56",  b  =  29°  37'  36". 

3.  c  =  41°  5'  6",  A  =  41°  32'  38". 

4.  a  =  12°  16'  42",  B=  79°  29 '45". 

5.  A  =  15°  38'  6",  B  =  80°  14'  41." 

6.  b  =  48°  27'  22",  c  =  56°  15'  43". 

7.  B  =  56°  15'  43",  c  =  58°  40'  13". 

8.  A  =  47°  37'  21",  B  =  61°  33'  4". 

9.  a  =  48°  27'  22",  c  =  64°  9'  43". 

10.  b  =  74°  21'  54",  A  =  38°  57'  12". 

11.  a  =  35°,  ^4  =  61°. 

12.  b  =  75°  45',  B  =  65°  38'. 

13.  b  =  105°  30',  c  =  80°  25'. 

14.  b  =  98°  35',  A  =  47°  38'. 

15.  b  =  79°  35',  B  =  80°  25'  20". 

56.  Quadrantal  triangles.  If  one  side  of  a  spherical 
triangle  is  a  quadrant,  the  angle  opposite  that  side  in 
the  polar  triangle  is  a  right  angle. 

From  the  two  given  parts  of  a  quadraiital  triangle  two 
parts  of  its  polar  triangle  may  be  obtained.  This  right 
triangle  may  then  be  solved  and,  from  these  solutions, 
the  unknown  parts  of  the  quadraiital  triangle  may  be 
obtained. 


On.  VII,  §  56]    RIGHT  AND  QUADRANTAL  TRIANGLES  127 

EXERCISE  XXXVI 

Find  the  remaining  parts  of  the  spherical  triangle  in  which 
c  =  90°  and 

1.  C=  163°  53'  38",  A  =  169°  29'  45". 

2.  C  =  141°  2'  48",  B  =  142°  5'  54". 

3.  B  =  140°  2'  56",  a  =  163°  53'  38". 

4.  A  =  148°  40'  13",  b  =  127°  54'  6". 

5.  a  =  138°  54'  54",  b  =  100°  30'  15". 

6.  a  =  65°,  A  =  48°  35'. 

7.  B  =  50°  38'  20",  b  =  75°  37'  30". 

8.  C  =  70°  30'  28",  b  =  128°  35'  12". 


CHAPTER   VIII 
OBLIQUE  TRIANGLES 

57.    Law  of  the  sines.  —  In  the  oblique  spherical  triangle 
ABC  draw  the  arc  CD  perpendicular  to  AB,  forming  the 
two  right  spherical  triangles  ADC 
and  CDB. 

Then  by  formula  [2], 

sin  DC  =  sin  a  sin  B, 

and  sin  DC  =  sin  b  sin  A. 

Hence 

sin  a  sin  B  =  sin  b  sin  A, 
sin  a     sin  A 


FIG.  7. 


or 


In  like  manner 
and 


sin  b     sin  />' 

sin  &  _  sin  B  t 
sin  c     sin  C 

sin  c_sin  C 
sin  a     sin  A 


In  this  proof  we  have  used  a  figure 
in  which  D  falls  between  A  and  B\ 
but  all  of  the  statements  will  be  seen 
to  be  true  for  Fig.  8,  in  which  D 
falls  without  AB,  if  we  note  that,  in 
this  case,  sin  DBC  =  sin  (TT  —  J5)  = 
sin  B. 

128 


CH.  VIII,  §  58] 


OBLIQUE   TRIANGLES 


129 


58.  Law  of  the  cosines.  —  Let  ABC  be  a  spherical  tri- 
angle in  which  two  of  the  sides,  b  and  <?,  are  each  less  than 
a  quadrant.  Join  its  vertices 
to  0,  the  centre  of  the  sphere. 
Through  D,  any  point  on  OA, 
pass  a  plane  DEF  perpen- 
dicular to  OA,  cutting  the  0- 
planes  OAO  and  OAB  in  the 
lines  DE  and  DF.  Then 
DE  and  DF  are  perpendicu-  FlG< 

lar  to  OA,  and  /.EDF  is  the  measure  of  the  spherical 
angle  A. 

In  the  triangle  OEF,  by  [31],  Pt.  I, 

EF*  =  OH2  +  OF*  -20E-  OF  cos  EOF. 


In  the  triangle  DEF,  by  [31],  Pt.  I, 

=  DE2  +  DF2  -  2  7X#  •  DF  cos 


Equating  these  two  values  of  ^/J?7  and  reducing,  we  have 
OE  -  OF  cos  EOF  =  01?  +  DE  -  DF  cos  EDF. 


or 


cos  a  =  cos  &  cose  +  sin  b  sine  cos  A. 


[12,  a 


In  the  above  proof  b  and  c  were  each  taken  less  than  a 
quadrant.  Let  the  student  show  by  a  method  similar  to 
that  used  in  Art.  53  that  the  formula  holds  for  all  values 
of  b  and  c. 


130  SPHERICAL  TRIGONOMETRY        [Cn.  VIII,  §  59 

By  interchanging  letters  in  the  cyclic  order  we  have 

cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  B9  [12,  £] 

cos  c  =  cos  a  cos  6  +  sin  a  sin  &  cos  C.  [12,  <?] 

Let  A' B'  C'  be  the  polar  triangle  of  ABC.  Since  the 
formulas  just  found  hold  for  all  triangles, 

cos  a1  =  cos  b'  cos  cf  +  sin  £>'  sin  c'  cos  ^4/, 
cos  bf  =  cos  £r  cos  ar  +  sin  <?'  sin  af  cos^r, 
cos  cr  =  cos  af  cos  6'  +  sin  ar  sin  5'  cos  Cf. 

But  a'  =  TT  —  A,  A'  =  TT  —  a,  etc.  Substituting  these 
values  and  noting  that  sin  (TT  —  #)  =  sin  rr,  and  cos  (TT  —  x) 
=  —  cos  ar,  we  have 

cos  ^1  =  -  cos  .B  cos  C  +  sin  JS  sin  C  cos  «,  [1 3,  a] 
cos  ^  =  -  cos  C  cos  ^1  +  sin  C  sin  ^1  cos  ft,  [13,  Z>] 
cos  C  =  -  cos  A  cos  B  +  sin  ^1  sin  B  cos  c.  [13,  c] 

59.  Formulas  [12]  and  [13]  are  unsuited  to  logarith- 
mic computation  ;  but  by  a  transformation  very  similar 
to  that  used  on  page  104  of  the  Plane  Trigonometry,  we 
may  obtain  from  them  formulas  which  are  well  adapted 
to  the  use  of  logarithms. 

From  [12,  a], 

cos  a  —  cos  b  cos  c 


cos  A  = 


sin  b  sin  c 


CH.  VIII,  §  59]  OBLIQUE   TRIANGLES  131 


But        sin  J  A  =  V"        ,  [231,  Pt.  1, 


sin  o  sin  c  —  cos  a  -}-  cos  o  cos  c 
2  sin  5  sin  c 


foogft-.Q-ooaa  fe     ^  pt   ^ 

»  V  ci  n   n  em   /» 


Z  Sill  0  Sill  g 


=     /sinj_(a_ 


—  <?)  sin  -j-  (^  —  5  +  g) 


sin  6  sin  g 

by  [29,  d],  Pt.  I. 

Let  a-f&  +  g=2s.     Then  i  (a  +  5  —  c)  =  s  —  c,  etc. 


.     T    ,         /sin  (s  —  c)  sin  (s  —  5) 
Hence        sm  A  ^L  =  \/—  — ^ L . 

*  cm   n  «m   /> 


sin  6  sin  <? 


Also          cos  \A  =  ^l  +  f7OS  A,  [24],  Pt.  I, 


Vcos  a  —  cos 
ty.  ein  /i  ei 


+  c) 


2  sin  6  sin  c 

by  [12]  and  [29,  •<*],  Pt.  I, 


sin  ( s  — 


sin  b  sin  g 
By  division, 


tan  I  A  = 


sin  s  sill  (s  —  a) 


-^ Ji 

dn  (s  —  a)  ^ 


/sin  (g  —  a)  sin  (g  —  6)  sin  (g  —  <?) 
sin  ( s  —  a )  *  sin  s 


132  SPHERICAL  TRIGONOMETRY        [Cn.  VIII,  §  60 


If  we  let         k  =  •>'i(''-«0»i»(«-a)''i"(«-'0 

sin  s 

tanH  =  ^(~^>  [H«] 

tanl*=s-nr^)>  t14'*] 

tan;c=^cf^)-  C14'<1 

Let  the  student  obtain  in  a  similar  manner  the  following 
formulas  from  formula  [13] : 

tan  |  a  =  K  cos  (8  -  A)9  [15,  a] 

tan|&  =  JKcos(S--B),  [15,5] 

tan  I  c  =  K  cos  (S  -  C),  [15,  c] 
in  which  2  $  =  A  +  B  +  C7,  and 


-  cos  8 


cos  (tf  -  A)  cos  (8  -  B)  cos  (#  -  (7) 

60.    Napier^s  analogies.  —  Dividing  [14,  a]  by  [14,  5], 

we  have 

sin  ^  A  cos  j  B  _  sin  (s  —  5) 


cos  JJ.  sin  !  .B      sin  (s—  a) 
By  composition  and  division, 


sin  j-  ^4.  cos  ^B  -\-  cos  j  J.  sin  \  B  _  sin  (s  —  b)  +  sin  (s  —  a) 
sin  J  A  cos  J  -S  —  cos  J  A  sin  J  j5      sin  (s  —  6)  —  sin  (s  —  «) 


Applying  [11],  [13],  [29,  a],  [29,5],  Pt.  I,  we  have 

[16] 


tanc 


tan^(a-ft) 


CH.  VIII,  §  60]  OBLIQUE   TRIANGLES  133 

Multiplying  [14,  a]  by  [14,  5],  we  have 

sin  I  A  sin  \  B  _  _  lfi_  _  _  sin  (s  —  c)  ^ 
cos  J  A  cos  ^  B     sin  («  —  a)  sin  (s  —  6)  sin  s 

By  inversion  ;  then  by  division  and  composition, 

cos  |  A  cos  j-  B  —  sin  j-  J.  sin  -j  J?  _  sin  s  —  sin  (s  —  c) 
cos  J  ^4.  cos  J  -B  +  sin  J  ^4.  sin  ^  5      sin  s  -f-  sin  (s  —  c) 

Applying  [12],  [14],  [29,  6],  [29,  a],  Pt.  I,  we  have 
cos|(^  +  B)^       ton|c 


Let  the  student  obtain  in  a  similar  manner  the  follow- 
ing formulas  from  [15,  a]  and  [15,  6]  : 


sin  £  (a  -b) 


cos  J  (a  +  6)  cot|c 


[18] 


[19] 


Let  the  student  also  obtain  from  each  of  the  above  four 
formulas  two  others  by  interchanging  the  letters. 

EXERCISE   XXXVII 

1.  Show  that  J  (a  +  &)  and  |-  (A  +  -B)  may  have  any  value 
less  than  180° ;   but  that  each  -of  the   other  angles  used  in 
Napier's  analogies  must  be  less  than  90°. 

2.  Show  by  the  aid  of  formulas  [17]  and  [19]  that  A  +  B  is 
less  than,  equal  to,  or  greater  than  180°,  according  as  a  -f-  b  is 
less  than,  equal  to,  or  .greater  than  180°;  and  the  converse. 

3.  Prove  that  in  any  spherical  triangle  each  angle  is  greater 
than  the  difference  between  180°  and  the  sum  of  the  other  two 
angles. 


134  SPHERICAL   TRIGONOMETRY        [On.  VIII,  §  61 

4.  Deduce  from  Napier's  Rules  for  the  right-angled  spheri- 
cal triangle  the  relation 

sin  a  _  sin  A 
sin  b      sin  B 

5.  Show  what  changes  occur  in  formulas  [11,  a,  &,  c],  (1) 
when  C  =  90°,  (2)  when  c  =  90°,  (3)  when  B=C=  90°. 

6.  If  a  perpendicular  be  dropped  from  vertex  Oof  an  oblique 
spherical  triangle  upon  the  opposite  side  AB,  to  meet  it  at  X, 
prove  by  Napier's  Rules  that,  disregarding  signs, 

sin  AX  _  cot  A 
sin  BX~  cotB' 

7.  By  means  of  Napier's  Rules,  derive  formulas  for  finding 
in  an  oblique  spherical  triangle  the  parts  required  in  the  follow- 
ing cases : 

(a)  Given  A,  C,  a,  required  b. 

(b)  Given  B,  C,  c,  required  b. 

8.  A  certain  point  X  on  a  sphere  is  joined  to  three  points 
P,  Q,  R  on  an  arc  of  a  great  circle.     By  application  of  the  law 
of  cosines  to  the  triangles  formed  and  by  reduction,  deduce  the 
formula,  sin  PQ  cos  RX+  sin  QR  cos  PX—  sin  PR  cos  QX=  0. 

61.  Solution  of  oblique  spherical  triangles.  —  When  any 
three  parts  of  a  spherical  triangle  are  given,  the  remaining 
parts  may  be  determined  by  the  aid  of  the  seven  formulas, 
[11]  and  [14]  to  [19]. 

Six  cases  may  occur.     There  may  be  given 

1.  the  three  sides, 

2.  the  three  angles, 

3.  two  sides  and  the  included  angle, 

4.  two  angles  and  the  included  side, 

5.  two  sides  and  the  angle  opposite  one  of  them, 

6.  two  angles  and  the  side  opposite  one  of  them. 


CH.  VIII,  §  62] 


OBLIQUE   TRIANGLES 


135 


There  will  be  one  determinate  solution  in  each  of  the 
first  four  cases,  but  in  5  and  6  there  may  be  two  solutions. 

62.    CASE  I.    Given  the  three  sides  cr,  6,  and  c. 
The  angles  may  be  determined  by  formula  [14]. 

tan  \A  = ,  tan  4  B  = — — , 

01 "  fat  —  a)  sm  (s  —  b) 


tan  1  0  =  - 


sin  (s 
k 


sn 


EXAMPLE  1.    Find    the  three  angles  of   the  spherical 
triangle  in  which 


a  =  16°  6'  22",  b  =  52°  5'  54",  and  c  =  61°  33'  4". 

s  =  64°  52'  40", 

log  sin  («—«)=  9.87627 
log  sin  (s  -  b)  =  9.34478 
log  sin  («-<?)=  8.76364 

27.98469 
=    9.95684 


s-  a  =  48° 46'  18", 
s-b  =12°  46' 46", 
s-c  =  3°  19' 36". 


log  k  =  9.01392 

log  sin  (s  -  a)  =  9.87627 
log  tan  J  A 


log  sin  s 
log  &2 
log& 
log& 


=  18.02785 
=    9.01392 

=  9.01392 
log  sin  (s-  5)  =  9.34478 
log  tan  ^  B 


=  9.13765        log  tan  1  B        =    9.66914 
38' 6".  £=500  2'  52". 

log  k  =    9.01392 

log  sin  (§-<?)=    8.76364 
log  tan  I  O        =  10.25028 

C  =121°  19'  46". 


136  SPHERICAL   TRIGONOMETRY        [Cn.  VIII,  §  63 

EXERCISE  XXXVIII 

Find  the  three  angles  of  the  spherical  triangle  in  which 

l.o=    22°  35' 52",     b=    40°   9' 21",  c=    56°  52' 23". 

2.  a=    35°  30' 24",     b=    38°  57' 12",  c=    56°  15' 43". 

3.  a=    39°  20' 24",     b=    41°   5'   6",  c=    60°  22' 24". 

4.  o=    41°  32' 38",     6=    44°  44' 17",  c=    57°  10'  4". 

5.  a=    52°   5' 54",     6=    61°  33'   4",  c=    83°  34' 56". 

6.  a-    56° 52' 23",     b=    80°  14' 41",  c  =  103°  59' 30". 

7.  a  =    68°12'58",     6=    80°14'41",  c  =  128°  11'  15". 

8.  a=    95°38'20",     6  =  108°26'30",  c=    56°27'48". 

9.  a  =  120°  22' 40",     6  =  111°  34'  27",  c=    96°  28' 35". 
10.    a=    56°  20' 20",     6=    56°  20' 20",  c=    60°  28' 38". 

63.    CASE  II.     Given  the  three  angles  A,  B,  and  C. 
The  sides  may  be  determined  by  formula  [15]. 

tan  1  a  =  Kcos  (tf  -  A),   tan  J  5  =  Jf  cos  (#  -  ^), 

tan  Jc=  Trccs^-  (7). 
The  logarithmic  work  is  very  similar  to  that  of  Case  I. 

EXERCISE  XXXIX 

Find  the  three  sides  of  the  spherical  triangle  in  which 

1.  ^1  =  50°    2' 56",   B  -56°  52'  23",    C=    86°  34' 33". 

2.  ^  =  47°  37' 21",   B  =  74°  18'  19",    C=    77°  48' 18". 

3.  .4  =  45°  26'  42",   B  =  47°  37'  21",    (7=  102°  16' 42". 

4.  A  =  15°  38'    6",   £  =  16°    6' 22",    C=  159°  44'  26". 


CH.  VIII,  §64]  OBLIQUE   TRIANGLES  137 

64.   CASE  III.     Given  two  sides  and  the  included  angle. 
The  two  remaining  angles  may  be  determined  by  for- 
mulas [18]  and  [19]. 

._.       sin  4  7  (a  —  b}      ,  i  n 
i 


tan  |-  (A  +  B)=  cot  *  °~ 

cos  I  (a  +  b) 

The  third  side  may  then  be  found  by  either  [11],  [16], 
or  [17]. 

EXAMPLE  1.  Find  the  remaining  parts  of  the  spherical 
triangle  in  which  a  =  40°  9;  21",  c  =  79°  29'  45",  B  = 
50°  2'  56". 

Here  we  must  use 

i  /-  n       A^      sin  -J  (<?  —  a)      ,  !  T> 
tan  4  (  O  —  A)  =  -;  —  f-7  --  ^  cot  J-  .B, 

sin  I  (c*  +  a) 


and 


cos       c  +  <«) 
i-  (c  _  «)  =  19°  40r  12",    and  J  (c  +  a)  =  59°  49'  33". 

log  sin  J  (<?-«>=   9.52712       log  cos  J  (<?-a)=   9.97389 
log  cot  J^=10.33084  log  cot  -1^=10.33084 

19.85796  20.30473 

logsinj(c  +  a)=   9.93677       log  cos  J  (<?  +  «)=   9.70125 


log  tan  -|(  (7-^)=   9.92119     log  tan  \  (C+A)  =  10.  60348 
i  (  (7-  A)  =  39°  49'  46"  £  (  C+  A)  =  76°  0'  28" 

Hence  A=    36°  10'  42", 

and  O=  115°  50'  14". 


138  SPHERICAL   TRIGONOMETRY        [CH.  VIII,  §65 

rr»     i   i         -7  •     7       sin  B  sin  a 

To  determine  0,  use  sin  6  =  -  — 

sin  .A 

log  sin  £=    9.88456 

log  sin  a  =    9.80947 

19.69403 

log  sin  A  =    9.77106 

log  sin  b   =    9.92297 

b  =  56°  52'  27". 

EXERCISE   XL 

Find  the  remaining  parts  of  the  spherical  triangle  in  which 
1.6  =  68°  12'  58",  c  =  80°  14'  41",  A  =  17°  20'  54". 

2.  a  =  27°  59'  4",  b  =  41°  5'  6",  C=  123°  44'  17". 

3.  a  =  29°  6'  11",  c  =  77°  43'  18",  B  =  38°  57'  12". 

4.  a  =  41°  5'  6",  &  =  60°  20'  54",  C=  77°  43'  18". 

5.  c  =  125°  20',  6  =  175°  36',  A  =  20°  28'  46". 

6.  c  =  98°  35'  26",  a  =  39°  48'  30",  B  =  47°  28'  42". 

7.  6  =  85°  35'  20",  c  =  73°  24'  26",  .4  -  95°  28'  40". 

8.  b  =  140°  38',  a  =  130°  28',  C=150°34'. 

65.    CASE  IV.     Given  two  angles  and  the  included  side. 

The  two  remaining  sides  may  be  determined  by  formulas 

[16]  and  [17]. 


The  third  angle  may  then  be  found  by  either  [10],  [18], 
or  [19]. 

The  logarithmic  work  is  very  similar  to  that  of  Case  III. 


CH.  VIII,  §66]  OBLIQUE  TRIANGLES  139 

EXERCISE   XLI 

Find  the  remaining  parts  of  the  spherical  triangle  in  which 

1.  c  =  107°  37'  55",  A  =  50°  2'  56",  B  =  64°  9'  43". 

2.  a  =  50°  2'  56",  £  =  61°  33'  4",  O  =  84°  53'  48". 

3.  b  =  56°  52'  23",  A  =  41°  32'  38",  C  =  111°  47'  4". 

4.  a  =  41°  5'  6",  £  =  56°  15'  43",  (7=  109°  45'  36". 

5.  A  =  48°  39'  20",  B  =  69°  28'  30",  c  =  58°  24'  36". 

6.  A  =  110°  48'  24",  C  =  60°  25'  48",  b  =  98°  59'  30". 
1.  B  =  98°  35'  28",  C=  99°  52'  48",  a  =  50°  50'  50". 
8.  A  =  110°  45'  38",  B  =  99°  37'  18",  c  =  120°  28'  20". 

66.    CASE  V.    Given  two  sides  and  the  angle  opposite  one 
of  them.  —  If  a,  5,  and  A  are  given,  B  may  be  found  from 

formula  fill.  .    ,    - 

.  sin  o  sin  A 


sin  -o  = 


sin  a 


If  sin  b  sin  ^4.  >  sin  a,  there  is  no  solution.  But  if 
sin  b  sin  A  <  sin  a,  there  are  sometimes  two  solutions. 
After  the  two  values  of  B  have  been  obtained,  the  number 
of  solutions  may  be  determined  from  the  fact  that  the 
greater  side  is  opposite  the  greater  angle.  It  will  also 
be  necessary  to  see  that  the  theorem  of  Problem  2,  Exer- 
cise xxxvii,  is  satisfied. 

The  remaining  parts,  c  and  C,  may  now  be  found  from 
formulas  [16]  and  [18],  or  from  [17]  and  [19]. 


140  SPHERICAL  TRIGONOMETRY        [Cn.  VIII,  §  06 

EXAMPLE  1.  Find  the  remaining  parts  of  tlie  spherical 
triangle  in  which  a  =  103°  10',  b  =  120°  12',  B  =  131°  40'. 

.      .,      sin  a  sin  B 

sin  A  =  -      —  -- 
sin  o 

log  sin  a  =    9.98843 

log  sin  B  =    9.87334 

19.86177 

log  sin  b  =    9.93665 
log  sin  -4=    9.92512 

A  =  57°  18'  45",  or  122°  41'  15". 

Both  of  these  values  of  A  will  be  seen  to  satisfy  the 
conditions  stated  above.  There  are,  therefore,  two  solu- 
tions. Using  the  first  value  of  A,  we  shall  proceed  to 
find  the  corresponding  values  of  c  and  0. 

J  (a  +  5)  =111°  41',  £(5-a)  =    8°  31', 

|-  (A  +  B)=    94°  29'  22",  \  (B-A)  =  37°  10'  37". 

Since  -|  (A  +  B)  is  near  90°,  more  accurate  results  will 
be  obtained  by  using  formulas  [17]  and  [19],  which  contain 
the  cosine.  «  »• 


cot  l  (7=  cos         +  a    tan  \  (B  +  X). 

cos  \(p  —  a) 


log  cos  %(B  +  A)=    8.89363  log  cos  J  (b  +  a)  =    9.56759 

log  tan  J  (5  4-  a)   =  10.40054  log  tan  |  (B  +  A}=  11.10504 

19^29417  20.67263 

log  cos  $(B-A)=    9.90133  log  cos  J  (5  -a)  =    9.99518 

log  tan  l  c  =    9.39284  log  cot  -J-  0  =  10.67745 

c  =  27°  45'  26".  C=  23°  44'  14". 


CH.  VIII,  §  67]  OBLIQUE   TRIANGLES  141 

In  each  of  the  above  formulas  two  factors  in  the  second 
member  are  negative.  The  first  members  are,  therefore, 
positive,  and  the  acute  values  of  |  c  and  ^  O  must  be 
chosen. 

Using  the  second  value  of  A  we  find,  by  the  aid  of  the 
same  formulas,  *=  113°  28' 14",  0=  127°  32'  56". 

EXERCISE  XLII 

Find  the  remaining  parts  of  the  spherical  triangle  in  which 

1.  a  =  16°  6'  22",  c  =  52°5'54",  A  =  15°  38'  7". 

2.  Z>  =  38°  57'  12",  c=  56°  15' 43",  £  =  47°  37' 21". 

3.  6  =  50°  2'  56",  c  =  56°  52'  23",  £  =  64°  9' 43". 
4.6  =  28°  35'  30",  c  =  30°  28'  15",  B  =  85°  38'  40". 

5.  S  =  86°9',  a  =  72°  18' 15",  6  =  71°  54' 15". 

6.  J.  =  120°35'28",  b  =  98°  48'  24",  a  =  105°  30'  30". 

7.  ^  =  103°  28' 12",  6  =  20°  25'  35",  a  =  28°  58' 25". 

Show  that  the  following  triangle  is  impossible ;  also  find  a 
value  for  c  such  that  B  shall  be  equal  to  90°. 

8.  0=98°  35' 28",  b  =  70°  35'  24",  c  =  50°  28'  22". 

67.  CASE  VI.  Given  two  angles  and  the  side  opposite 
one  of  them. 

If    J.,    5,    and   a   are   given,    b   may   be   found    from 

formula   fill.  .     n   - 

.    7      sin  U  sin  a 

sin  0  = : — 

sin  JL 

The  number  of  solutions  may  be  determined  as  in  Case  V. 

The  remaining  parts  c  and  O  may  now  be  found  as  in 

Case  5,  by  formulas  [16]  and  [18],  or  by  [17]  and  [19 j. 

The  logarithmic  work  is  very  similar  to  that  of  Case  V. 


142  SPHERICAL  TRIGONOMETRY        [Cn.  VIII,  §  67 

EXERCISE  XLIII 
Find  the  remaining  parts  of  the  spherical  triangle  in  which 

1.  b  =  56°  15'  43",  £  =  38°  57'  12",  C=  138°  54'  54". 

2.  b  =  48°  20',  A  =  76°  50',  B  =  59°  48'. 

3.  A  =  70°  30'  28",  a  =  45°  28'  32",  B  =  60°  20'  32'-'. 

4.  .4  =  78°  47'  20",  a  =  63°  49'  10",  C  =  80°  25'  30". 

5.  c  =  112°  49'  24",  (7=  152°  49'  27.5",  A  =  29°  42'  13.7". 

6.  C=S°  48'  48",  c  =  85°  26'  45",  5  =  23°  49'  15". 

7.  A  =  57°  48'  23",  B  =  120°  38'  27",  a  =  48°  25'  20". 

8.  J.  =  70°28',  a  =  80°  25'  40",  C=125°28'. 

REVIEW  EXERCISE 

1.  If  a  median  be  drawn  from  the  vertex  C  of  a  spherical 
triangle  to  the  opposite  side  c,  and  the  parts  of  the  angle 
adjacent  to  sides  a  and  b  of  the  triangle  be  named  a  and  ft 

respectively,  show  that  5^  = 


sin  ft      sin  a 

2.  The  city  of  Quito  is  situated  nearly  on  the  equator  and 
its  longitude  is  78°  50'  west  of  Greenwich.     The  latitude  of 
Greenwich  is  51°  28'.     Find  the  distance  from  Greenwich  to 
Quito  (on  the  arc  of  a  great  circle).     Assume  the  radius  of  the 
earth  to  be  4000  miles. 

3.  In  a  spherical  triangle  whose  sides  are  48°,*  57°,  and  65°, 
respectively,  a  median  is  drawn  to  the  side  whose  length  is  48° 
from  the  opposite  vertex.     Find  the  length  of  the  median  and 
also  the  parts  into  which  it  divides  the  angle. 

4.  If  the  values  of  two  sides  of  a  spherical  triangle  are  « 
and  ft  and  the  angle  between  them  n,  find  the  length  of  the 
perpendicular  upon  the  third  side   from   the   vertex  of  the 
angle  n. 


CH.  VIII,  §  67]  OBLIQUE   TRIANGLES  143 

5.  If   a  lune  whose  angle  is  a  be  drawn  upon  a  sphere 
whose  radius  is  c  feet,  and  an  arc  of  a  great  circle  be  drawn  to 
intersect  at  equal  angles  the  sides  of  the  lune,  making  the  part 
of  the  arc  so  included  /?,  find  the  distance  from  each  vertex 
of  the  lune  to  the  transverse  arc  in  feet. 

6.  A  ship  starting  from  a  point  on  the  equator  in  longitude 
130°  West  sailed  for  3  days,  arriving  at  a  point  whose  latitude 
is  20°  North  and  longitude  150°  West.     What  was  its  rate  per 
hour,  allowing  the  radius  of  the  earth  to  be  4000  miles  ? 

7.  The  sides  of  a  spherical  triangle  enclosing  an  angle 
of  75°  are  respectively  60°  and  54°.     Find  the  length  of  the 
bisector  of  the  angle  and  the  angles  it  makes  with  the  base. 

8.  There   is   a   regular    tetrahedron   each   of   whose   face 
angles  is  60°.     Find  the  angle  between  any  two  faces. 

NOTE.  —  Suppose  one  vertex  of  the  tetrahedron  to  be  at  the  centre 
of  a  sphere  whose  radius  is  an  edge  of  the  tetrahedron.  The  other 
three  vertices  of  tho  solid  will  determine  upon  the  surface  of  the  sphere 
the  vertices  of  a  spherical  triangle  whose  sides  are  measured  by  the 
face  angles  of  the  tetrahedron. 

9.  If  the  face  angle  at  the  vertex  of  a  regular  four-sided 
pyramid  is  50°,  find  the  angle  between  any  two  lateral  faces. 

10.  Find  the  area  of  a  spherical  triangle  whose  sides  are 
45°  26',  53°  44',  68°  46',  respectively,  on  a  sphere  whose  radius 
is  10  ft. 

NOTE._  The  formula  for  the  area  of  a  spherical  triangle  is: 
area  =  """  ^  ,  where  E  denotes  the  excess  in  degrees  of  the  sum  of 

-LoU 

the  angles  of  the  triangle  over  180°.     This  excess  may  be  found  when 
the  three  sides  of  the  triangle  are  given  by  THuilier's  Formula, 


tan  I E  =  Vtan  \  s  tan  \  (s  -  a)  tan  \  (.s  -  b)  tan  £(*  -  c)> 

in  which  a,  &,  and  c  denote  the  sides  of  the  triangle,  and  s,  as  usual, 
the  half  sum  of  the  sides. 


144  SPHERICAL  TRIGONOMETRY         [Cn.  VIII,  §  67 

11.  In  a  sphere  of  radius  12  is  a  spherical  pyramid  whose 
base  is  a  spherical  triangle  of  which  the  sides  are  85°,  65°, 
and  120°.     The  vertex  of  the  pyramid  being  at  the  centre  of 
the  sphere,  find  its  volume. 

12.  If  a  line  makes  an  angle  0  with  its  projection  on  a 
plane  passing  through  one  end  of  the  line  and  if  the  projec- 
tion makes  an  angle  <£  with  a  second  line  drawn  in  a  plane 
which  interse'cts  the  first  plane  in  the  line  of  the  projection  at 
an  angle  of  30°,  show  that  the  angle  between  the  first  line  and 
the  second  is  cos"1  \  (2  cos  0  cos  <£  sin  0  sin  </>). 

13.  A  flight  of  stone  steps  faces  due  south.     A  rod  rests 
with  one  end  on  a  step  and  leans  against  the  edge  of  the  step 
above,  in  a  plane  perpendicular   to  the  steps.     At  noon  the 
horizontal  part  of  the  shadow  is  marked  on  the  step  and  also 
the  vertical  part.     If  the  rod  makes  an  angle  0  with  the  step 
upon  which  its  foot  rests,  show  that  t  hours  after  noon  the 
angle  the  horizontal  part  of  the  shadow  makes  with  its  posi- 
tion  at   noon    may   be    determined    by   the    equation    tan   x 
=  sin  6  tan  t,  and  the  angle  the  vertical  part  of  the  shadow 
makes    with    its    position    at    noon,   by   the    equation   tan?/ 
=  cos  0  tan  t. 

NOTE.  —  This  example  illustrates  the  principle  of  both  the  hori- 
zontal and  the  vertical  sun  dial.  6  represents  the  latitude  of  the 
place.  Let  the  lower  end  of  the  rod  be  the  centre  of  a  sphere  whose 
surface  is  pierced  by  the  rod  and  its  two  horizontal  shadows  in  three 
points  which  are  the  vertices  of  a  spherical  right  triangle.  By  means 
of  this  triangle  the  first  relation  may  be  proved. 

It  should  be  remembered  that  each  hour  of  time  corresponds  to  15°. 

14.  If  the  longitude  of  New  York  is  40°  43',  find  the  angles 
which  the  shadows  the  sun  would  cast  at  three  o'clock  P.M. 
upon  a  dial  constructed  according  to  the  principle  of  Ex.  13, 
would  make  with  the  shadows  cast  at  noon. 


ANSWERS 


ANSWERS 


Exercise  I,  page  4 

1.   (a)  sin  A  =  ^,    cos  A  =  if,    tan  A  =  T\,   sec  A  =  Jjj,    esc  J.  =  J/, 
cot  .4  =  -^,  vers  .4  =  T^,  covers  A  =  T8^. 

(6)  sin  -4  =  |,  cos  A  =  f ,  tan  ^4  —  f ,  sec  A  =  £,  esc  J.  =  |,  cot  A  =  f . 

(c)  sin  ^4  =  T8r,     cos  A  =  Jf ,     tan  ^4  =  T%,      sec  J.  =  J£,      esc  ^4  —  y , 
cot  ^4  =  -1/. 

(d)  sin  -4  =  |,   cos  ^4  =  |  V5,   tan  ^1  =i  f  V5,    sec  ^1  =  f  V5,  esc  ^1  =  f , 
cot  A=  JV6. 

7.  c  =  16.  8.    ^1  =  20°.  9.   ^1  =  40°,  a  =  5.0346,    c  =  7.83. 

10.   A  =  20°,  ^  =  70°,  c  =  266.  11.  B  =  20°,  c  =  29.24,  a  =  27.475. 

12.   .4  =  70°,  a  =  18.794,  b  =  6.84. 

Exercise  II,  page  6 

1.  cos  20°,  sin  5°,  esc  27°,  cot  33°  12',  tan  4'. 

2.  x  =  45°.  3.    x  =  30°.  4.   x  =  15°. 

Exercise  III,  page  8 
1.    J.  2.   5.  3.    f.  4.    9.  5.    1.  6.    \. 

Exercise  IV,  page  12 

1.  cos  A  =  I \/7,  tan  .4  =  f  V7,  sec  J.  =  f  V7,  esc  ^4  =  f ,  cot  ^1  =  £ \/7. 

2.  sin  ^  =  |  V6,  tan  ^1  =  2  \/6,  sec  A  =  5,  esc  ^1  =  &  V6,  cot  .4  =  &  V6. 

3.  sin -4  =  ^yVlO,     cos  ^4  =  3^  VlO,      sec^l=VlO,      csc^l  =  ^VlO, 
cot  -4  =  |. 

4.  sin  J_=  ^V  A/37,       cos  A  =  &  A/37,       tan  -4  =  |,       sec  JL  =  J  V3T, 
esc  A  =  V3T. 

5.  See  Art.  4. 

147 


148 

6. 

cot  A 

7. 
8. 
9. 

cot  A 


ANSWERS 

sin  A  =  rV,      cos  A  =  &  VTT,     tan  A  =  -fa  VTT,      sec  A  =  £§  VTT, 


cos  A  =  j\,  tan  A  =  *£,  sec  A  =  -1/,  esc  -4  =  |f,  cot  .4.  = 
sin  A  =  $,  tan  .4  =  f ,  sec  A  —  f ,  esc  ^4  =  £,  cot  ^4  =  f . 
sin  4  = 


,  cos  A  =  -,    tan  .4  =  v  a2  —  1,   esc  A  = 


14. 


15. 


(a) 
00 

fa*\ 

cos  .4 
tan  A 

cot  A 
sec  .4 
esc  A 
sin  J. 

cos  -4 

cot  A 
sec  .4 

sin7  .4 

=  Vl  -  sin2  A, 
sin  .4 

(6)    sin^l  = 

Vl  -cos2  .4, 

Vl  -  cos2  A 

Vl  -  sin2  .4 

cos  A 
cos  A 

Vl      sin2  .4 

_           1  

Vl  -  cos2  A 
1 

esc  A  — 

1 

1 

sin  A 

tan  A 

(cZ)   sin  A  •= 
cos  A  — 

Vl  -  cos2  A 

Vsec2  A  —  1 

Vl  +  tan2  A 
1 

sec  .4 

1 

Vl  4-  tan2  A 

1 

tan^l  = 

cot  A  = 

esc  A  = 
4-  sin2  .4-1 

sec  A 

Vsec2  A  —  1, 
1 
Vsec2  A  -  1 
sec  A 

tan  A 

=  Vl  +  tan2  .4, 

Vl  4-  tan2  A 

tan  A 

-2sin5,44-2sm3.4 

Vsec2  A-  I 

sin2  ^1- sin4 
1  -  sin2  A  -  sin3  A 
1  -  sin2  A 


00 


16.    (a)  cos^lVl-cos2A     (6) 

00 


cos 


1  —  sin2  A  4-  sin4  A 
1  -sin2  A 

,,    1  -  3  cos2  A  4  -°>  cos4  A 

cos4  ^4  -  cos2  A 
1  l-tan^l  +  tnn'M 


Vl  4-  tan2  A    1 4-  tan  J.  4  tan-  A 


(tan  A  -  1)  Vl  +  tan2 .4 
tan  .4 


ANSWERS 


149 


Exercise  VI,  page  18 


3.  .4=36°  52',  7?=53°   8',  c=5. 

4.  ^1=12°  41',  5=77°  19',  c=41. 

5.  ,4=67°  23',  J?=22°37',  6  =  5. 


7.  -4=81°  12',  S=  8°  48',  a=84. 

8.  #=55°18',  6  =  17.83,  c  =  21.08. 

9.  ^1=66°  26',  a  =  13.75,  6  =  5.997. 


6.  .4=79°  37',  5=10°  23',  6  =  11.      10.  5=76°   8',  a=4.197,  c=17.51. 


Exercise  VII,  page  19 


3.  ^  =  39°  54' 28",  c  =  8850.6. 

4.  A  =  41°  48'  35",  6  =  2484.3. 

5.  .4  =  56°  26'  27",  6  =  0.3015. 

6.  6  =  10322,  c  =  11287. 

7.  a  =0.778,  6  =  0.4036. 

8.  6  =  454.43,  c  =  499. 

9.  a  =  2.005,  6  =  1.287. 

10.  A  =  53°  15'  6",  c  =  2194. 

11.  ^1  =  2°  27' 52",  c  =  13.48. 


12.  a  =  1760.5,  c  =  1762.2. 

13.  ^1  =  53°  7'  48",  a  =  11.2. 

14.  ^1  =  78°  20' 39",  c  =  811.74. 

15.  a  =  518.61,  6  =  161.95. 

16.  6  =24.187,  c  =  24.23. 

17.  A  =  43°  44'  51",  6  =  0.00679. 

18.  a  =  761.17,  6  =  76.42. 

19.  a  =  965.93,  6  =  258.82. 

20.  .4  =  71°  38',  6  =  0.334. 


1.  31°  45' 33". 

2.  67.4ft. 

3.  36°  52' 12". 

4.  166.43ft. 

5.  23.3ft. 

6.  202.2ft. 


Exercise  VIII,  page  20 

7.  34°  54' 54",  13.  40.98ft. 
72°  32'  33",   72°  32'  33".  14.  140.88  ft. 

8.  61.6ft.  15.  44°  25' 37". 

9.  838.8  ft.  16.  9.81  ft. 

10.  660yd.  19.  169.3ft. 

11.  3°  13' 29".  20.  769.8ft. 

12.  75ft. 


.      Exercise  IX,  page  29 

1.    ^TT,  !„-,  2^  |r>  n^  !„..  2.    36°,  20°,  120°,  150°,  900°. 


3.    J£*yf7r,  15  TT. 


4.    2. 


Exercise  XII,  page  46 

1.   cos  x  =  ±  f  V6~,      tan  cc  =  ±  TV  VG~>     sec  x  =  ±  r52  V6,     esc  x  =  —  5, 


2.  sin  a;  =  -  f  V2,       tan  x  =  -  2  \/2,       sec  x  =  3,        cscx  =  -  f\/2", 
cotse  =-  JV2. 

3.  sinx  =  -j\VTo,      cosx  =  fjyVIO,     secx=VlO, 
cot  x  =  -    . 


150  ANSWERS 


4.  sinx  =  ±  iVl5,       coso:  =  |,  tana;  =  ±  Vl5,       cscx  =±  j^VlS, 
cotx  =  ±  ^VTs. 

5.  sin^l  =  ±^   cos^  =  x/x2~y2,    tan^  =  ± ^ ,    csc^  =  ±?, 

x  Vx2  -  va                      ^ 


Exercise  XIV,  page  55 

1.  |V3,  f;  -0.868,4.924;   -4.698,   -1.71. 

2.  5,  -  53°  8';  or  -  5,  126°  52'.       .      3.    197°  27'  28",  or  -  17°  27'  28". 

4.  If  the  triangle  is  described  in  the  positive  direction  of  rotation,  the 
angles  are  120°,   -  120°,   -  120°,  5  ;  -  5  ;  -  5. 

5.  5V3;  -5V3;  0;  7£  ;  -  7J. 

Exercise  XV,  page  63 

2.   i(V6->/2).       3.  2>V3._   4.  -*-(  V6  -  V2),  i(V6  +  V2),  2-  V~3. 
5.   i  V3  +  i  V2  ;   -  J-  Vl  -  |V2. 

8.  cos  a  cos  j3  cos  7  +  cos  a  sin  /3  sin  7—  sin  a  sin  /8  cos  7+  sin  a  cos  j8  sin  7. 
g      tan  ct  —  tan  ft  +  tan  7  +  tan  a  tan  ft  tan  7 
1  +  tan  «  tan  /3  —  tan  a  tan  7  +  tan  /3  tan  7 

Exercise  XVI,  page  65 
1.    ±iV3.  2.   -f.  3.    3.43. 


5.  cos3<*  =  4cos3<*-3cos  a.   tan  3  g  =  3  tan  "  ~  . 

1-3  tan2 

6.  sin  4  ct  =  8  sin  ex,  cos3  «  —  4  sin  a  cos  a. 
cos  4  a  =  8  cos4  a  -  8  cos2  a  +  1. 
tan4<*=    4tan«-4tan«tt    m 

1-6  tan2  a  +  tan4  a 

7.  sin  5  a  =  5  sin  a  —  20  sin3  a  +  16  sin5  «, 
cos  5  a  =  5  cos  a  —  20  cos3  a  +  16  cos5  a. 


Exercise  XVII,  page  68 

1.    0.316,  0.9487  ;  0.78. 


2.    sin22°30'=^v2_V2,    cos  22°  30'  =2+  V2,    tan22°30f  =  V2-1. 
4.    sin  165°  =J(v/0-v/2),  cos  165°=  -J(V2+  V6),  tan  165°=  V3-2. 


ANSWERS  151 

Exercise  XX,  page  75 

1.    mr.  2.     O-fi)7r.  3.    nir.  4.    (2n  +  ^)ir. 

5.  (2rc-|)7r.  6.    2  nir.  7.   (2n+l)7r.  8.    2  n?r. 
9.     (2n+l)ir.                              10.    O  +  ^)TT.                              11.    (n  +  f)ir. 

12.    (2n-i)ir,   (2n-f)ir.  13.    (2n  +  £)ir,   (2  n  +  f  )TT. 

14.    (2n±i>.  15.   (2n±J>.  16.   (2n±|)r. 

Exercise  XXII,  page  81 

1.  45°,  135°,  225°,  315°  ;  mr  ±  -. 

4 

2.  60°,  90°,  120°,  270°;  2  nir  ±-,  2  nir +§,  (2n  +  l)7r-£; 

2  o  3 

3.  45°,  225° ;  mr  +  -. 

4.  0°,  60°,  300°  ;  2  nir,  2  mr  ±£ •  5.    0°,  180°;  WTT. 

o 

6.  15°,  75°,  195°,  255°;  mr  +  -^  mr  +  ~ 

7.  60°,  180°,  300° ;  2  mr  ±  -,  (2  n  +  1)  TT. 

o 

8.  45°,  135°,  225°,  315° ;  mr  ±  -. 

4 

9.  45°,  165°  58',  225°,  345°  58' ;  mr  +  -,  mr  +  tan-i(-  £). 

10.    30°,  60°,  120°,  150°,  210°,  240°,  300°,  330°;  mr  ±  -,  nir  ±-. 

6  3 

Exercise  XXIII,  page  83 

1.  45°,  225° ;  mr  +  -•  3.    285°,  345° ;  2  mr  -  — ,  2  WTT  -  — . 

2.  0°,  90° ;  2  WTT,  2  WTT  +  -.        4.    120° ;  2  WTT  -f — . 

2  3 

5.  24°  27',  261°  49' ;  n  •  360°  -  36°  52'  ±  61°  19'. 

6.  27°  58',  135°,  242°  2',  315°;  mr  +— ,  Jsin-1(2V2  -  2). 

7.  0°,  90°,  180°;  nir,  2wir+£. 

8.  0°,  90°;  2  nir,  2  nir +2:. 

9.  30°,  270° ;  2  WTT  +  -,  2  WTT  -  -• 

6  2 

10.    46°  24',  90° ;  2  mr  +  -,  n  •  360°  +  46°  24'.  ' 
2 


152 


ANSWERS 


Exercise  XXIV,  page  86 

1.  90°,  270°;  nir+f- 

2.  51°  19',  180°,  308°  41';  (2  n  +  !)JT,  cos-if.  3.   0°,  180°;  mr. 

4.  0°,  60°,  120°,  180°,  240°,  300°;  mr,  (n  +  |)TT  ±  -• 

6 

5.  0°,     7£°,     37i°,     971°,     127|°,     180°,    etc.;     mr, 


6.    30°,  60°,  90°,  120°,  150°,  210°,  240°,  270°,  300°,  330°; 


7.  45°,  60°,  120°,  135°,  225°,  240°,  300°,  315°;  mr±-,  mr  ±-. 

8.  0°,  180°;  mr. 

9.  18°,  162°,  234°,  306°;  sin-i  ~  1  ±  V5. 

10.  18°,  54°,  90°,  126°,  162°,  etc. ; 

11.  45°,  90°,  135°,  225°,  270°,  315° ;  mr  ±  -,  mr  +  -. 

4  2 

12.  22°  6',  67°  54' ;  £  sin-i  |  (5  -  Vl3). 

13.  0°,  65°  4',  252°  45' ;  2  mr, 


14.  45°,  67|°,  90°,  1571°,  225°,  247|°,  270°,  337|°;  mr  +  1, 

W7T         37T 

T  +  ~F' 

15.  22°  30',  112°  30'  ;  £    nir  +  -    . 


16.    60°,  90°,  120°,  240°,  270°,  300°  ;  mr  ±  £,  WTT  +  ^. 

3  2 


Exercise  XXV,  page  87 

1.   f.         5.  £\/6.         7.  -2  -V3.         9.  i.          11.  f.          13. 
14.    0°,  45°,  180°,  225°  ;  nir,  mr  +  ^. 

18.    22|°,  112  J°,  202|°,  292^°;  \  I  mr  +-Y 


23.    (1) 


(np  +  n  +  p  -  1)  (np  -  n  -  p  -  1) 
sec  A  sec  5 


;  (2) 


esc  A  esc  B 


csc2.B-l- 
27.    18°,  54°,  126°,  162°,  198°,  234°,  306°,  342°;  sin-i  ±  £(V5  ±  1). 


ANSWERS  153 

Exercise  XXVI,  page  96 

2.  C  =  54°  18',  6  =  3.317,  c  =  3.925. 

3.  A  =  123°  12',  a  =  23.63,  c  =  20.51. 

4.  B  =25°  12',  c  =  227.7,  6  =  157.4. 

5.  C  =  35°  4',  b  =  577.3,  c  =  468.9. 

7.  O  =  87°  32'  5",  6  =  17.632,  c  =  21.746. 

8.  ^1  =  38°  21'  47",  a  =  13.509,  b  =  17.632. 

9.  A  =  29°  25'  18",  b  =  2675.9,  c  =  3674. 

10.  B  =  67°  27'  33",  b  =  77.08,  c  =  79.06. 

11.  B  -  100°  22'  45",  a  =  1337.2,  6  =  1758.9. 

12.  ^4  =  139°  21' 42",  a  =  100,  c  =  63.15. 

Exercise  XXVII,  page  97 
2.   c  =  8.9.  3.    a  =  13.  4.    c  =  2.  5.    b  =  V3. 

Exercise  XXVIII,  page  99 

2.  A  =  12°  22',  B  =  149°  15',  c  =  34.37. 

3.  A  =  64°  19'  28",  B  =  42°  24'  22",  c  =  612.06. 

4.  A  =  84°  12'  33",   C  =  45°  46'  59",  b  =  0.5591. 

5.  #  =  37°  48' 5",   C  =  42°  11'  55",  a  =  0.0117. 

6.  ^  =  33°  5' 18",   Cr=41°0'42",  6  =  96.42. 

7.  4  =  31°  50'  20",  .B  =  50°  4'  25",  c  =  3139.9. 

8.  A  =  133°  51'  34",  5=11°  59'  10",  c  =  2479.2. 

9.  A  =  70°  22'  38",  B  =  21°  24'  42",  c  =  33.787. 
10.  .4  =  72°  40'  41",  #  =  15°8'1",  c  =  15.272. 

Exercise  XXIX,  page  102 

2.  A  =  41°  13'  0",   C  =  87°  37'  54",  c  =  116.82. 

3.  4  =  11°  26'  58",   C  =  84°  16'  31",  c  =  401. 

4.  B  =  48°  27'  20",   C  =  90°,  b  =  360. 

5.  A  =  46°  52'  10",   C  =  111°  53'  25",  c  =  883.65. 
A1  =  133°  7'  50",   C'  =  25°  37'  45",  fe  =  411.92. 

6.  A  =  29°  11' 39",   0  =  91° 34' 21",  c  =  8.853. 

7.  ^1  =  83°  40',   C=71°5'47",  c  =  670.1. 
4'  =  96°  20',   C"  =  58°  25'  47",  c'  =  603.5. 

8.  -4  =  19°  19'  3",   (7  =142°  59' 48",  c  =  89.15. 
4'  =  160°  40'  57",   C"  =  1°  37'  54",  c'  =  4.218. 

9.  4  =  10°  54'  58",   C  =132°  12',  c  =  946.68. 
10.   A  =  57°  37'  18",  C  =  90°,  a  =  88. 


154  ANSWERS 

Exercise  XXX,  page  103 

2.  .4  =  60°,  #  =  32°  12',   C7  =  87°48'. 

3.  A  =  56°  7',  .#  —  81°  47',   C  =  42°  6'. 

4.  .4  =  56°  15',  5  =  59°  51',   Cy  =  63°54'. 

5.  ^  =  38°  57',  £  =  47°  41',   C  =93°  22'. 

Exercise  XXXI,  page  106 

2.  ^1  =  7°  37' 42",  B  =  61°  55'  38",   C  =  110°  26 '40". 

3.  ^1  =  47°  38',  _B  =  68°1'6",   C  =  64°  20'  54". 

4.  A  =  85°  55'  7",  5  =  43°  57'  33",   C  =  50°  7'  20". 

5.  J.  =  23°  32'  12",  B  =  56°  8'  42",    C  =  100°  19'  6". 

6.  .4  =  59°  39'  30",  #  =  42°  35'  20",   C  =  77°  45'  10". 

7.  A  =  33°  15'  39",  B  =  50°  56',    C=  95°  48'  21". 

8.  ^  =  37°  22' 19",  B  =  38°  15'  41",    C  =104°  22'. 

9.  A  -  36°  45'  14",  B  =  53°  3'  8",   C  =  90°  11'  38". 

Exercise  XXXII,  page  108 

1.  14.68.  3.    0.00815.  5.    0.156.  7.    28621.  9.    0.0265. 

2.  1259.6.          4.    88.66.  6.    2520.  8.    1285.3.          10.    63.34. 

Exercise  XXXIII,  page  109 

1.  27.65,  80.08  ;  65°  19' 58",  14.    75.13  ft.  ;  225.4  ft. 

114°  40' 2".  15.  11646ft. 

2.  169.45ft.  16.  Z(cot|8-cota). 

3.  4.43ft.;  7.35ft.  17.  753.1yd. 

4.  114.41  ft.  18.  91.772  ft. 

5.  46.14  ft.  ;  99.92  ft.  19.  47.168  ft.  ;  16°  20'. 

6.  171.08yd.  20.  8. 574  miles. 

7.  13  miles  per  hour,  nearly.  21.  12.15ft. 

8.  114.6  in.  23.  20  V3  ft. 

9.  2000  (2  V3- 3)  yd.  26.  106ft. 

10.  23.87  ft. ;  15.18  ft.  27.  5(2  +  V3)  ft. 

11.  39.97  ft.  ;  29.99  ft.  .        28.  9.24ft. 

12.  520.44  ft.  29.  180  ft.  ;  500  ft. 

13.  330.72ft.  ;  138°  35' 30".  31.  15.14ft. 

32.  tan-1      tan  e  tan  ^       ;  tan-1  (tan  0  cot  0  sec  0) ; 

Vtan2  0  -f  tan2  0 


tan-1  (tan  0  cot  0  sec  0) ;  tan"1  (cot  0  cot  0  Vtan2  0  +  tan2  0) . 

35.  50V6ft.  3g  I  .     2Zsin2« 

36.  250V3ft. 


ANSWERS  155 

Exercise  XXXIV,  page  118 

8.    100°,  105°,  75°;  157.08,  164.934,  117.81.        •  9.    110°,  100°,  70°. 
10.   00°  ;  a  great  circle  and  its  pole. 

Exercise  XXXV,  page  126 

1.  c  =15°  38' 6",  .4  =  38°  57 '12",  J3  =  52°5'54". 

2.  c=40°9'21",  A  =  47°  37'  21",  .B  =  50°  2'  56". 

3.  a=  25°  50'  17",  b  =  33°  7'  37",  B  =  56°  15'  43". 

4.  &  =  48°  54'  54",  c  =  50°  2'  56",  A  =  16°  6'  22". 

5.  a  =  12°  16'  42",  &  =  51°  2'  48",  c  =  52°  5'  54".    . 

6.  a  =  33°  7'  37",  A  =  41°  5'  6",  B  =  64°  9'  43". 

7.  a  =  42°  22'  39",  6  =  45°  15'  43",  .4  =  52°  5' 54". 

8.  a  =  39°  57'  4",  &  =  49°  50'  39",  c  =  60°  22'  24" . 

9.  6  =  48°  54'  54",  A  =  56°  15'  43",  B  =  56°  52'  23". 

10.  a  =  37°  54'  6",  c  =  77°  43' 18",  B  =  80°  14'  41". 

11.  c=  40°  58'  50",  6  =  22°  50' 19",  .S  =  36°  17'  17", 
c  =  139°  1'  10'  ,  6=  157°  9'  41",  B  =  143°  42'  43". 

12.  Impossible. 

13.  A  =  127°  30'  11",  a  =  128°  32',  B  =  102°  14'  30". 

14.  5  =  96°  19' 51.6",  c  =  95°  48  '28",  a  =  47°  18'  44". 

15.  a  =  66°  37',  c  =  85°  52',  ^1  =  66°  57' 48". 
a  =  113°  23' ,  c  =  94°  8',  A  =  113°  2'  12". 

Exercise  XXXVI,  page  127 

1.  B=  167°  43'  18",  a  =  138°  54'  54",  b  =  129°  57'  4". 

2.  A=  170°  14'  41",  a  =  164°  21'  54",  &  =  102°  16'  42". 

3.  C  =  138°  54'  54",  A  =  169°  29'  45",  &  =  102°  16'  42". 

4.  C=  138°  15'  43",  B  =  146°  15'  43",  a  =  132°  22'  39". 

5.  C=  102°  16'  42",  A  =  140°  2'  56",    B  =  106°  6'  22". 

6.  B  =  31°  54'  40",  C  =  55°  50'  7",  ft  =  39°  42'  23". 
B  =  148°  5'  20",  C  =  124°  9'  53",  &  =  140°  17'  37". 

7.  A  =  18°  12'  24",   C  =  52°  57'  12",  a  =  23°  2'  44". 

A  =  161°  47'  36",  C  =  127°  2'  48",  a  =  156°  57'  16". 

8.  B  =  132°  31'  45",  A  =  60°  25'  39",  a  =  67°  18'  38". 

Exercise  XXXVIII,  page  136 

1.  A  =  20°  35'  37",  B  =  36°  10'  39",   C  -  129°  57'  4". 

2.  A  =  43°  2'  7",  5  =  47°  37'  21",   C  =  102°  16'  42".    • 

3.  .A  =  45°  26'  42",  B  =  47°  37'  21",   C  =  102°  16'  42". 

4.  A  -  52°  6(  54",  B  =  56°  52'  23",   C  =  88°  42'  27". 


156  ANSWERS 

5.  A  =  50°  2'  56",  B  =  58°  40'  13",   C  =  105°  6'  41". 

6.  A  =  52°  5'  54",  B  =  68°  12'  58",   C  =  113°  53'  57". 

7.  A  =  52°  5'  54",  B  =  56°  52'  23",   C  =  138°  5'  42". 

8.  A  =  84°  26'  48",  B  =  108°  24'  54",   C  =  56°  28'  32". 

9.  A  =  126°  18'  42",  B  =  119°  42'  8",   C  =  111°  51'  42". 
10.  A  =  67°  9'  28",  B  =  67°  9'  28",  C  =  74°  27'  56". 

Exercise  XXXIX,  page  136 

1.  a  =  36°  10'  39",  b  =  40°  9'  21",  c  =  50°  13'  58". 

2.  a  =  38°  57'  12",  b  =  55D  1'  2",  c  =  56°  15'  43". 

3.  a  =  39°  20'  24",  b  =  41°  5'  6",  c  =  60°  22'  24". 

4.  a  =  50°  2'  56",  6  =  52°  5'  54",  c  =  99°  57'  42". 

Exercise  XL,  page  138 

l.o  =  20°  32'  33",  B  =  52°  5'  54",   C  =  123°  7'  37". 

2.  c  =  60°  22'  24",  J.  =  26°  40'  20",  B  =  38°  57'  12". 

3.  6  =  56°  15'  43",  A  =  21°  34'  28",   C  =  132°  22'  39". 

4.  c  =  60°  22'  24",  ^  =  47°  37'  21",  B  =  77°  39'  26". 

5.  C  =  21°  41'  25",  B  =  178°  0'  29",  a  =  50°  33'  38". 

6.  C  =  129°  53'  4",  ,4  =  29°  47'  28",  6  =  71°  45'  15". 

7.  B  =  84°  11'  45",  C  =  72°  59'  41",  a  =  93°  58'  18". 

8.  B  =  161°  46'  32",  A  =  157°  58'  8",  c  =  85°  19'  46". 

Exercise  XLI,  page  139 

1.  a  =  56°  52'  23",  b  =  19°  29'  45",   C=  119°  15'  56". 

2.  b  =  52°  5'  54",  c  =  63°  21'  53'',  A  =  58° 40'  13". 

3.  a  =  44°  44'  17",  c  =  80°14'4i",  £  =  52°  5'  54". 

4.  b  =  60°  22'  24",  c  =  79°  39'  38",  A  =  38°  57'  12". 

5.  a  =  40°  12'  34",  b  =  53°  38'  28",   C=82°9'. 
6. '  o  =  112°  25'  37",  c  =  59°  19'  25",  B  =  87°  14'. 

7.  b  =  108°  20'  51",  c  =  108°  58'  5",  A  =  53°  53'  6". 

8.  o  =  108°  32'  10",  b  =  88°  35'  18",   C  =  121°  47'  14", 

Exercise  XLII,  page  141 

1.  6  =  40°  32'  33",  B  =  39°  9'  35",   (7  =  129°  57' 4". 
6  =  61°  33'  4",  B  =  121°  19'  47",   C  =  50°  2'  56". 

2.  a  =  35°  30'  24",  A  =  43°  2'  7",   (7  =  102°  16'  42". 
a  =  55°  1'  2",  A  =  74°  18'  19",   C  =  77°  43'  18". 

3.  a  =  21°  7'  35",  A  =  25°  26'  16",   O  =  100°  30'  15' . 
a  =  46°  059",  4  =  47°  39' 0",   C  =  79°  29'  25". 


ANSWERS  157 

4.  Impossible. 

5.  A  =  90°,  C  =  12°  29'  4",  c  =  11°  53'  42". 

6.  B  =  61°  59',  c  =  13°  38'  16",  C  =  12°  9'  24  '. 
B  =  118°  1',  c  =  132°  29'  46",  C  .=  138°  48'  6". 

7.  -B  =  44°"28'  46",  c  =  16°  35'  58",  C  =  34°  59'  24". 

8.  c  =  68°  50'  36". 

Exercise  XLIII,  page  142 

1.  a  =  5°  21'  59",  c  =  60°  22'  24",  A  =  4°  3'  15". 

a  =  77°  43'  18",  c  =  119°  37'  37",  A  =  47°  37'  21". 

2.  A=  57°  18'  43",   C  =  66°  31'  42",  c  =  52°  27'  4". 

^1  =  122°  41'  17",   C  =  152°  14'  42",  c  =  156°  15'  54". 

3.  b  =  41°  5'  17",  c  =  42°  55'  48",   C  -  64°  14'. 

4.  c  =  64°  26'  20",  6  =.  40°  48'  50",  B  =  45°  35'  50". 

c  =.  115°  33'  40",  b  =  176°  34'  16",  B  =  176°  15'  4"  , 

5.  a  =  90°,  &  =  25°  57'  12",  B  =  12°  28'  38". 

6.  Impossible. 

7.  &  =  49°  30'  48",  c  =  2°  5'  26",   (7=2°  21'  54". 

b  =  130°  29'  12",  c  =  118°  5'  56",   C=  94°  4'  4". 

8.  c  =  121°  33'  9",  b  =  77°  39'  20",  B  =  69°  0'  18", 

Review  Exercise 

2.  5799.8  miles. 

3.  median  =  58°  3'  15". 

angles  =  26°  12'  29"  and  28°  31'  6". 

_  sin  n  tan  a  _  \ 
Vsin2  n  +  (tan  a  -  tan  ft  cos  n)2  J* 
tan  £-  cot 


5.  . 

180 

6.  27  miles  (nearly). 

7.  bisector  =  50°  35'  21". 

angles  =  84°  34'  and  95°  26'. 

8.  70°  31'  43".  10.    42.43ft. 

9.  87°  15'  2".  11.    904.7808. 
14.    33°  7'  2";  37°  9'  37". 


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